Mixing
For complex vectors $z_1$ and $z_2$, How do I show that if $|z_1+z_2|=|z_1|+|z_2|$ then the vectors $z_1$ and...
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In my complex analysis class, we went over a geometric proof of this with the triangle inequality, but I'm trying to find a more algebraic proof. I'm also trying not to use Arg because we haven't gone over it in my class yet and all I really know about an argument is that if $z_1$, $z_2$, and 0 are collinear than they'd all have the same angle off the real axis.
I haven't really gotten particularly far. I can convert the absolute values (based on the definition of modulus that, for a complex number $z=x+iy$, $|z|=sqrt{(x^2+y^2)}$). I can then square both sides of the equation twice and then simplify to get something that looks really easy to work with, but I'm not sure what the results of that could tell me about how the three points are collinear.
Any help would be much appreciated.
complex-analysis
asked Jan 29 at 21:34
MasieMasie
283
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In my complex analysis class, we went over a geometric proof of this with the triangle inequality, but I'm trying to find a more algebraic proof. I'm also trying not to use Arg because we haven't gone over it in my class yet and all I really know about an argument is that if $z_1$, $z_2$, and 0 are collinear than they'd all have the same angle off the real axis.
I haven't really gotten particularly far. I can convert the absolute values (based on the definition of modulus that, for a complex number $z=x+iy$, $|z|=sqrt{(x^2+y^2)}$). I can then square both sides of the equation twice and then simplify to get something that looks really easy to work with, but I'm not sure what the results of that could tell me about how the three points are collinear.
Any help would be much appreciated.
complex-analysis
asked Jan 29 at 21:34
MasieMasie
283
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Actually antiparallel vectors won't satisfy the equation (unless you consider 0 and any complex number to be antiparallel).
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– Daniel Schepler
Jan 29 at 22:03
add a comment |
$begingroup$
In my complex analysis class, we went over a geometric proof of this with the triangle inequality, but I'm trying to find a more algebraic proof. I'm also trying not to use Arg because we haven't gone over it in my class yet and all I really know about an argument is that if $z_1$, $z_2$, and 0 are collinear than they'd all have the same angle off the real axis.
I haven't really gotten particularly far. I can convert the absolute values (based on the definition of modulus that, for a complex number $z=x+iy$, $|z|=sqrt{(x^2+y^2)}$). I can then square both sides of the equation twice and then simplify to get something that looks really easy to work with, but I'm not sure what the results of that could tell me about how the three points are collinear.
Any help would be much appreciated.
complex-analysis
asked Jan 29 at 21:34
MasieMasie
283
$endgroup$
In my complex analysis class, we went over a geometric proof of this with the triangle inequality, but I'm trying to find a more algebraic proof. I'm also trying not to use Arg because we haven't gone over it in my class yet and all I really know about an argument is that if $z_1$, $z_2$, and 0 are collinear than they'd all have the same angle off the real axis.
I haven't really gotten particularly far. I can convert the absolute values (based on the definition of modulus that, for a complex number $z=x+iy$, $|z|=sqrt{(x^2+y^2)}$). I can then square both sides of the equation twice and then simplify to get something that looks really easy to work with, but I'm not sure what the results of that could tell me about how the three points are collinear.
Any help would be much appreciated.
complex-analysis
complex-analysis
asked Jan 29 at 21:34
MasieMasie
283
asked Jan 29 at 21:34
MasieMasie
283
edited Jan 29 at 21:41
Masie
asked Jan 29 at 21:34
MasieMasie
283
asked Jan 29 at 21:34
MasieMasie
283
asked Jan 29 at 21:34
MasieMasie
283
283
$begingroup$
Actually antiparallel vectors won't satisfy the equation (unless you consider 0 and any complex number to be antiparallel).
$endgroup$
– Daniel Schepler
Jan 29 at 22:03
add a comment |
$begingroup$
Actually antiparallel vectors won't satisfy the equation (unless you consider 0 and any complex number to be antiparallel).
$endgroup$
– Daniel Schepler
Jan 29 at 22:03
$begingroup$
Actually antiparallel vectors won't satisfy the equation (unless you consider 0 and any complex number to be antiparallel).
$endgroup$
– Daniel Schepler
Jan 29 at 22:03
$begingroup$
Actually antiparallel vectors won't satisfy the equation (unless you consider 0 and any complex number to be antiparallel).
$endgroup$
– Daniel Schepler
Jan 29 at 22:03
add a comment |
5 Answers
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Say $z_1 = a+ib$ and $z_2 = c+id$, which correspond to the vectors $(a,b)$ and $(c,d)$ in $mathbb{R}^2$. Their norms squared are $lvert z_1 rvert^2 = a^2+b^2$ and $lvert z_2 rvert^2=c^2+d^2$. Now $$z_1+z_2 = a+c+i(b+d),$$ hence its norm squared is $$lvert z_1+z_2rvert^2 = (a+c)^2+(b+d)^2 = lvert z_1rvert^2 + lvert z_2 rvert^2+2(ac+bd).$$
Moreover
$$
lvert z_1z_2rvert = sqrt{(a^2+b^2)(c^2+d^2)}.
$$
Now, your equation is equivalent to
$$lvert z_1+z_2rvert^2 = left(lvert z_1rvert+lvert z_2rvertright)^2=lvert z_1rvert^2+lvert z_2rvert^2+2lvert z_1z_2rvert$$
which implies
$$ac+bd=sqrt{(a^2+b^2)(c^2+d^2)},$$
namely $(ad-bc)^2=0$. So $ad-bc=0$, i.e. $z_1$ is parallel to $z_2$. You can see this as follows: if $langle cdot, cdot rangle$ denotes the standard scalar product in $mathbb{R}^2$, then $ad-bc=0$ is the same as $langle (a,b), (d,-c)rangle=0$. But $langle (d,-c),(c,d)rangle = 0$, which means that $z_2$ is orthogonal to $(d,-c)$, which is in turn orthogonal to $z_1$. So finally $z_1$ and $z_2$ are parallel.
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
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Thank you so much!
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– Masie
Jan 30 at 2:47
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If $z_1$ is a positive real number, what does this look like?
$$|z_1+(x_2+y_2i)|=|z_1|+|x_2+y_2i|=z_1+sqrt{x_2^2+y_2^2}$$
$$|z_1+(x_2+y_2i)| = sqrt{(z_1+x_2)^2+y_2^2}$$
Equate the two, and square them:
$$z_1^2+x_2^2+y_2^2+2z_1sqrt{x_2^2+y_2^2} = left(z_1+sqrt{x_2^2+y_2^2}right)^2 = (z_1+x_2)^2+y_2^2 = z_1^2+x_2^2+y_2^2 + 2z_1x_2$$
$$2z_1sqrt{x_2^2+y_2^2}=2z_1x_2$$
So then, $x_2=sqrt{x_2^2+y_2^2}ge |x_2|$. Equality there is only possible if $y_2=0$, and equality of $x_2$ and $|x_2|$ is then only possible if $x_2ge 0$. So that's it - for positive real $z_1$, $|z_1+z_2|=|z_1|+|z_2|$ only if $z_2$ is a nonnegative real number. Either one of them's zero, or they're parallel.
Of course, we would like something that works for general $z$. For that, consider what happens if we multiply both $z_1$ and $z_2$ by the same unit vector $w$: $|z_1w|=|z_1|cdot |w|=z_1$ and $|z_1w+z_2w|=|(z_1+z_2)w|=|z_1+z_2|cdot |w|=|z_1+z_2|$. None of the absolute values change, so we can multiply a solution by $w$ to get another solution. In particular, if $|z_1+z_2|=|z_1|+|z_2|$ (and they're nonzero), then by taking $w=frac{|z_1|}{z_1}$, we get a new solution $z_1',z_2'$ with $z_1'=|z_1|$ a positive real number. By our previous work, this makes $z_2'$ also positive real. Undo the rotation: $z_1=frac1wcdot z_1'$ and $z_2=frac1wcdot z_2'$ are both positive real multiples of the same unit vector $frac1w$, so they're parallel. Victory - equality in the triangle inequality means they're parallel.
OK, there's also the case when one or both of the vectors is zero. Let's just say that zero is parallel to everything, and be done with that.
What about the "antiparallel" in your statement? That has to do with equalities like $|z_1+z_2|=|z_1|-|z_2|$ or $|z_1+z_2|=|z_2|-|z_1|$.
answered Jan 29 at 22:05
jmerryjmerry
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Let’s assume
$$z_1=x_1+iy_1$$
$$z_2=x_2+iy_2$$
Computing one gets
$$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=x_1^2+y_1^2+x_2^2+y_2^2+2(x_1x_2+y_1y_2)$$
And
$$(|z_1|+|z_2|)^2=x_1^2+y_1^2+x_2^2+y_2^2+2|z_1z_2|$$
The identity we are investigating is therefore equivalent to
$$|z_1z_2|=(x_1x_2+y_1y_2)$$
Equivalently
$$|z_1z_2|^2=(x_1x_2+y_1y_2)^2$$
Now $z_1z_2=x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)$ which means that
$$|z_1z_2|^2=(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2$$
After expanding and identifying, one gets
$$(x_1y_2)^2+(x_2y_1)^2-2x_1x_2y_1y_2=(x_1y_2-x_2y_1)^2=0$$
And this tells us that $x_1y_2-x_2y_1=0$ which is equivalent to $z_1$ and $z_2$ are colinear
answered Jan 29 at 22:10
marwalixmarwalix
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I will answer the question in the heading (not including $0$).
Let $z_k=x_k+iy_k$ Square both sides and remove common terms to get: $(x_1+iy_1)(x_2-iy_2)+(x_1-iy_1)(x_2+iy_2)=(2(x_1x_2+y_1y_2))=2sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$. Now square both sides and eliminate common terms to get $2x_1x_2y_1y_2=x_1^2y_2^2+x_2^2y_1^2$ Assume none of the terms $=0$, divide by the product and get $2=frac{x_1y_2}{x_2y_1}+frac{x_2y_1}{x_1y_2}$ Notice that the two terms on the right are reciprocal, su that with $u=$ one ratio, we have a quadratic $u^2-2u+1=0$, so that the ratio $=1$, forcing the vectors to be scalar multiples of each other.
answered Jan 29 at 22:21
herb steinbergherb steinberg
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Let $z_1=r_1e^{itheta_1}=r_1costheta_1+ir_1sintheta_1, z_2=r_2e^{itheta_2}=r_2costheta_2+ir_2sintheta_2$.
Then $mid z_1+z_2mid=mid (r_1costheta_1 +r_2costheta_2)+i(r_1sintheta_1 +r_2sintheta_2)mid=sqrt{r_1^2cos^2theta_1 +2r_1r_2costheta_1costheta_2+r_2^2cos^2theta_2+r_1^2sin^2theta_1
+2r_1r_2sintheta_1 sintheta_2 +r_2^2sin^2theta_2}=sqrt{r_1^2+r_2^2+2r_1r_2(costheta_1, sintheta_1) cdot (costheta_2,sintheta_2) }=r_1+r_2implies (costheta_1, sintheta_1) cdot (costheta_2, sintheta_2)=1implies theta_1 =pmtheta_2 $.
Now I know you said you didn't want to use "arg", but all I used was polar coordinates. And the statement is really about the arguments of the two numbers.
answered Jan 29 at 22:27
Chris CusterChris Custer
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I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand.
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– Masie
Jan 30 at 2:48
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Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand.
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– Chris Custer
Jan 30 at 3:07
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5 Answers
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5 Answers
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Say $z_1 = a+ib$ and $z_2 = c+id$, which correspond to the vectors $(a,b)$ and $(c,d)$ in $mathbb{R}^2$. Their norms squared are $lvert z_1 rvert^2 = a^2+b^2$ and $lvert z_2 rvert^2=c^2+d^2$. Now $$z_1+z_2 = a+c+i(b+d),$$ hence its norm squared is $$lvert z_1+z_2rvert^2 = (a+c)^2+(b+d)^2 = lvert z_1rvert^2 + lvert z_2 rvert^2+2(ac+bd).$$
Moreover
$$
lvert z_1z_2rvert = sqrt{(a^2+b^2)(c^2+d^2)}.
$$
Now, your equation is equivalent to
$$lvert z_1+z_2rvert^2 = left(lvert z_1rvert+lvert z_2rvertright)^2=lvert z_1rvert^2+lvert z_2rvert^2+2lvert z_1z_2rvert$$
which implies
$$ac+bd=sqrt{(a^2+b^2)(c^2+d^2)},$$
namely $(ad-bc)^2=0$. So $ad-bc=0$, i.e. $z_1$ is parallel to $z_2$. You can see this as follows: if $langle cdot, cdot rangle$ denotes the standard scalar product in $mathbb{R}^2$, then $ad-bc=0$ is the same as $langle (a,b), (d,-c)rangle=0$. But $langle (d,-c),(c,d)rangle = 0$, which means that $z_2$ is orthogonal to $(d,-c)$, which is in turn orthogonal to $z_1$. So finally $z_1$ and $z_2$ are parallel.
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
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Thank you so much!
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– Masie
Jan 30 at 2:47
add a comment |
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Say $z_1 = a+ib$ and $z_2 = c+id$, which correspond to the vectors $(a,b)$ and $(c,d)$ in $mathbb{R}^2$. Their norms squared are $lvert z_1 rvert^2 = a^2+b^2$ and $lvert z_2 rvert^2=c^2+d^2$. Now $$z_1+z_2 = a+c+i(b+d),$$ hence its norm squared is $$lvert z_1+z_2rvert^2 = (a+c)^2+(b+d)^2 = lvert z_1rvert^2 + lvert z_2 rvert^2+2(ac+bd).$$
Moreover
$$
lvert z_1z_2rvert = sqrt{(a^2+b^2)(c^2+d^2)}.
$$
Now, your equation is equivalent to
$$lvert z_1+z_2rvert^2 = left(lvert z_1rvert+lvert z_2rvertright)^2=lvert z_1rvert^2+lvert z_2rvert^2+2lvert z_1z_2rvert$$
which implies
$$ac+bd=sqrt{(a^2+b^2)(c^2+d^2)},$$
namely $(ad-bc)^2=0$. So $ad-bc=0$, i.e. $z_1$ is parallel to $z_2$. You can see this as follows: if $langle cdot, cdot rangle$ denotes the standard scalar product in $mathbb{R}^2$, then $ad-bc=0$ is the same as $langle (a,b), (d,-c)rangle=0$. But $langle (d,-c),(c,d)rangle = 0$, which means that $z_2$ is orthogonal to $(d,-c)$, which is in turn orthogonal to $z_1$. So finally $z_1$ and $z_2$ are parallel.
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
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Thank you so much!
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– Masie
Jan 30 at 2:47
add a comment |
$begingroup$
Say $z_1 = a+ib$ and $z_2 = c+id$, which correspond to the vectors $(a,b)$ and $(c,d)$ in $mathbb{R}^2$. Their norms squared are $lvert z_1 rvert^2 = a^2+b^2$ and $lvert z_2 rvert^2=c^2+d^2$. Now $$z_1+z_2 = a+c+i(b+d),$$ hence its norm squared is $$lvert z_1+z_2rvert^2 = (a+c)^2+(b+d)^2 = lvert z_1rvert^2 + lvert z_2 rvert^2+2(ac+bd).$$
Moreover
$$
lvert z_1z_2rvert = sqrt{(a^2+b^2)(c^2+d^2)}.
$$
Now, your equation is equivalent to
$$lvert z_1+z_2rvert^2 = left(lvert z_1rvert+lvert z_2rvertright)^2=lvert z_1rvert^2+lvert z_2rvert^2+2lvert z_1z_2rvert$$
which implies
$$ac+bd=sqrt{(a^2+b^2)(c^2+d^2)},$$
namely $(ad-bc)^2=0$. So $ad-bc=0$, i.e. $z_1$ is parallel to $z_2$. You can see this as follows: if $langle cdot, cdot rangle$ denotes the standard scalar product in $mathbb{R}^2$, then $ad-bc=0$ is the same as $langle (a,b), (d,-c)rangle=0$. But $langle (d,-c),(c,d)rangle = 0$, which means that $z_2$ is orthogonal to $(d,-c)$, which is in turn orthogonal to $z_1$. So finally $z_1$ and $z_2$ are parallel.
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
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Say $z_1 = a+ib$ and $z_2 = c+id$, which correspond to the vectors $(a,b)$ and $(c,d)$ in $mathbb{R}^2$. Their norms squared are $lvert z_1 rvert^2 = a^2+b^2$ and $lvert z_2 rvert^2=c^2+d^2$. Now $$z_1+z_2 = a+c+i(b+d),$$ hence its norm squared is $$lvert z_1+z_2rvert^2 = (a+c)^2+(b+d)^2 = lvert z_1rvert^2 + lvert z_2 rvert^2+2(ac+bd).$$
Moreover
$$
lvert z_1z_2rvert = sqrt{(a^2+b^2)(c^2+d^2)}.
$$
Now, your equation is equivalent to
$$lvert z_1+z_2rvert^2 = left(lvert z_1rvert+lvert z_2rvertright)^2=lvert z_1rvert^2+lvert z_2rvert^2+2lvert z_1z_2rvert$$
which implies
$$ac+bd=sqrt{(a^2+b^2)(c^2+d^2)},$$
namely $(ad-bc)^2=0$. So $ad-bc=0$, i.e. $z_1$ is parallel to $z_2$. You can see this as follows: if $langle cdot, cdot rangle$ denotes the standard scalar product in $mathbb{R}^2$, then $ad-bc=0$ is the same as $langle (a,b), (d,-c)rangle=0$. But $langle (d,-c),(c,d)rangle = 0$, which means that $z_2$ is orthogonal to $(d,-c)$, which is in turn orthogonal to $z_1$. So finally $z_1$ and $z_2$ are parallel.
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
edited Jan 29 at 22:00
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
answered Jan 29 at 21:41
GibbsGibbs
5,4383927
5,4383927
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Thank you so much!
$endgroup$
– Masie
Jan 30 at 2:47
add a comment |
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Thank you so much!
$endgroup$
– Masie
Jan 30 at 2:47
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Thank you so much!
$endgroup$
– Masie
Jan 30 at 2:47
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Thank you so much!
$endgroup$
– Masie
Jan 30 at 2:47
add a comment |
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If $z_1$ is a positive real number, what does this look like?
$$|z_1+(x_2+y_2i)|=|z_1|+|x_2+y_2i|=z_1+sqrt{x_2^2+y_2^2}$$
$$|z_1+(x_2+y_2i)| = sqrt{(z_1+x_2)^2+y_2^2}$$
Equate the two, and square them:
$$z_1^2+x_2^2+y_2^2+2z_1sqrt{x_2^2+y_2^2} = left(z_1+sqrt{x_2^2+y_2^2}right)^2 = (z_1+x_2)^2+y_2^2 = z_1^2+x_2^2+y_2^2 + 2z_1x_2$$
$$2z_1sqrt{x_2^2+y_2^2}=2z_1x_2$$
So then, $x_2=sqrt{x_2^2+y_2^2}ge |x_2|$. Equality there is only possible if $y_2=0$, and equality of $x_2$ and $|x_2|$ is then only possible if $x_2ge 0$. So that's it - for positive real $z_1$, $|z_1+z_2|=|z_1|+|z_2|$ only if $z_2$ is a nonnegative real number. Either one of them's zero, or they're parallel.
Of course, we would like something that works for general $z$. For that, consider what happens if we multiply both $z_1$ and $z_2$ by the same unit vector $w$: $|z_1w|=|z_1|cdot |w|=z_1$ and $|z_1w+z_2w|=|(z_1+z_2)w|=|z_1+z_2|cdot |w|=|z_1+z_2|$. None of the absolute values change, so we can multiply a solution by $w$ to get another solution. In particular, if $|z_1+z_2|=|z_1|+|z_2|$ (and they're nonzero), then by taking $w=frac{|z_1|}{z_1}$, we get a new solution $z_1',z_2'$ with $z_1'=|z_1|$ a positive real number. By our previous work, this makes $z_2'$ also positive real. Undo the rotation: $z_1=frac1wcdot z_1'$ and $z_2=frac1wcdot z_2'$ are both positive real multiples of the same unit vector $frac1w$, so they're parallel. Victory - equality in the triangle inequality means they're parallel.
OK, there's also the case when one or both of the vectors is zero. Let's just say that zero is parallel to everything, and be done with that.
What about the "antiparallel" in your statement? That has to do with equalities like $|z_1+z_2|=|z_1|-|z_2|$ or $|z_1+z_2|=|z_2|-|z_1|$.
answered Jan 29 at 22:05
jmerryjmerry
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If $z_1$ is a positive real number, what does this look like?
$$|z_1+(x_2+y_2i)|=|z_1|+|x_2+y_2i|=z_1+sqrt{x_2^2+y_2^2}$$
$$|z_1+(x_2+y_2i)| = sqrt{(z_1+x_2)^2+y_2^2}$$
Equate the two, and square them:
$$z_1^2+x_2^2+y_2^2+2z_1sqrt{x_2^2+y_2^2} = left(z_1+sqrt{x_2^2+y_2^2}right)^2 = (z_1+x_2)^2+y_2^2 = z_1^2+x_2^2+y_2^2 + 2z_1x_2$$
$$2z_1sqrt{x_2^2+y_2^2}=2z_1x_2$$
So then, $x_2=sqrt{x_2^2+y_2^2}ge |x_2|$. Equality there is only possible if $y_2=0$, and equality of $x_2$ and $|x_2|$ is then only possible if $x_2ge 0$. So that's it - for positive real $z_1$, $|z_1+z_2|=|z_1|+|z_2|$ only if $z_2$ is a nonnegative real number. Either one of them's zero, or they're parallel.
Of course, we would like something that works for general $z$. For that, consider what happens if we multiply both $z_1$ and $z_2$ by the same unit vector $w$: $|z_1w|=|z_1|cdot |w|=z_1$ and $|z_1w+z_2w|=|(z_1+z_2)w|=|z_1+z_2|cdot |w|=|z_1+z_2|$. None of the absolute values change, so we can multiply a solution by $w$ to get another solution. In particular, if $|z_1+z_2|=|z_1|+|z_2|$ (and they're nonzero), then by taking $w=frac{|z_1|}{z_1}$, we get a new solution $z_1',z_2'$ with $z_1'=|z_1|$ a positive real number. By our previous work, this makes $z_2'$ also positive real. Undo the rotation: $z_1=frac1wcdot z_1'$ and $z_2=frac1wcdot z_2'$ are both positive real multiples of the same unit vector $frac1w$, so they're parallel. Victory - equality in the triangle inequality means they're parallel.
OK, there's also the case when one or both of the vectors is zero. Let's just say that zero is parallel to everything, and be done with that.
What about the "antiparallel" in your statement? That has to do with equalities like $|z_1+z_2|=|z_1|-|z_2|$ or $|z_1+z_2|=|z_2|-|z_1|$.
answered Jan 29 at 22:05
jmerryjmerry
16.9k11633
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add a comment |
$begingroup$
If $z_1$ is a positive real number, what does this look like?
$$|z_1+(x_2+y_2i)|=|z_1|+|x_2+y_2i|=z_1+sqrt{x_2^2+y_2^2}$$
$$|z_1+(x_2+y_2i)| = sqrt{(z_1+x_2)^2+y_2^2}$$
Equate the two, and square them:
$$z_1^2+x_2^2+y_2^2+2z_1sqrt{x_2^2+y_2^2} = left(z_1+sqrt{x_2^2+y_2^2}right)^2 = (z_1+x_2)^2+y_2^2 = z_1^2+x_2^2+y_2^2 + 2z_1x_2$$
$$2z_1sqrt{x_2^2+y_2^2}=2z_1x_2$$
So then, $x_2=sqrt{x_2^2+y_2^2}ge |x_2|$. Equality there is only possible if $y_2=0$, and equality of $x_2$ and $|x_2|$ is then only possible if $x_2ge 0$. So that's it - for positive real $z_1$, $|z_1+z_2|=|z_1|+|z_2|$ only if $z_2$ is a nonnegative real number. Either one of them's zero, or they're parallel.
Of course, we would like something that works for general $z$. For that, consider what happens if we multiply both $z_1$ and $z_2$ by the same unit vector $w$: $|z_1w|=|z_1|cdot |w|=z_1$ and $|z_1w+z_2w|=|(z_1+z_2)w|=|z_1+z_2|cdot |w|=|z_1+z_2|$. None of the absolute values change, so we can multiply a solution by $w$ to get another solution. In particular, if $|z_1+z_2|=|z_1|+|z_2|$ (and they're nonzero), then by taking $w=frac{|z_1|}{z_1}$, we get a new solution $z_1',z_2'$ with $z_1'=|z_1|$ a positive real number. By our previous work, this makes $z_2'$ also positive real. Undo the rotation: $z_1=frac1wcdot z_1'$ and $z_2=frac1wcdot z_2'$ are both positive real multiples of the same unit vector $frac1w$, so they're parallel. Victory - equality in the triangle inequality means they're parallel.
OK, there's also the case when one or both of the vectors is zero. Let's just say that zero is parallel to everything, and be done with that.
What about the "antiparallel" in your statement? That has to do with equalities like $|z_1+z_2|=|z_1|-|z_2|$ or $|z_1+z_2|=|z_2|-|z_1|$.
answered Jan 29 at 22:05
jmerryjmerry
16.9k11633
$endgroup$
If $z_1$ is a positive real number, what does this look like?
$$|z_1+(x_2+y_2i)|=|z_1|+|x_2+y_2i|=z_1+sqrt{x_2^2+y_2^2}$$
$$|z_1+(x_2+y_2i)| = sqrt{(z_1+x_2)^2+y_2^2}$$
Equate the two, and square them:
$$z_1^2+x_2^2+y_2^2+2z_1sqrt{x_2^2+y_2^2} = left(z_1+sqrt{x_2^2+y_2^2}right)^2 = (z_1+x_2)^2+y_2^2 = z_1^2+x_2^2+y_2^2 + 2z_1x_2$$
$$2z_1sqrt{x_2^2+y_2^2}=2z_1x_2$$
So then, $x_2=sqrt{x_2^2+y_2^2}ge |x_2|$. Equality there is only possible if $y_2=0$, and equality of $x_2$ and $|x_2|$ is then only possible if $x_2ge 0$. So that's it - for positive real $z_1$, $|z_1+z_2|=|z_1|+|z_2|$ only if $z_2$ is a nonnegative real number. Either one of them's zero, or they're parallel.
Of course, we would like something that works for general $z$. For that, consider what happens if we multiply both $z_1$ and $z_2$ by the same unit vector $w$: $|z_1w|=|z_1|cdot |w|=z_1$ and $|z_1w+z_2w|=|(z_1+z_2)w|=|z_1+z_2|cdot |w|=|z_1+z_2|$. None of the absolute values change, so we can multiply a solution by $w$ to get another solution. In particular, if $|z_1+z_2|=|z_1|+|z_2|$ (and they're nonzero), then by taking $w=frac{|z_1|}{z_1}$, we get a new solution $z_1',z_2'$ with $z_1'=|z_1|$ a positive real number. By our previous work, this makes $z_2'$ also positive real. Undo the rotation: $z_1=frac1wcdot z_1'$ and $z_2=frac1wcdot z_2'$ are both positive real multiples of the same unit vector $frac1w$, so they're parallel. Victory - equality in the triangle inequality means they're parallel.
OK, there's also the case when one or both of the vectors is zero. Let's just say that zero is parallel to everything, and be done with that.
What about the "antiparallel" in your statement? That has to do with equalities like $|z_1+z_2|=|z_1|-|z_2|$ or $|z_1+z_2|=|z_2|-|z_1|$.
answered Jan 29 at 22:05
jmerryjmerry
16.9k11633
answered Jan 29 at 22:05
jmerryjmerry
16.9k11633
answered Jan 29 at 22:05
jmerryjmerry
16.9k11633
answered Jan 29 at 22:05
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
$begingroup$
Let’s assume
$$z_1=x_1+iy_1$$
$$z_2=x_2+iy_2$$
Computing one gets
$$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=x_1^2+y_1^2+x_2^2+y_2^2+2(x_1x_2+y_1y_2)$$
And
$$(|z_1|+|z_2|)^2=x_1^2+y_1^2+x_2^2+y_2^2+2|z_1z_2|$$
The identity we are investigating is therefore equivalent to
$$|z_1z_2|=(x_1x_2+y_1y_2)$$
Equivalently
$$|z_1z_2|^2=(x_1x_2+y_1y_2)^2$$
Now $z_1z_2=x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)$ which means that
$$|z_1z_2|^2=(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2$$
After expanding and identifying, one gets
$$(x_1y_2)^2+(x_2y_1)^2-2x_1x_2y_1y_2=(x_1y_2-x_2y_1)^2=0$$
And this tells us that $x_1y_2-x_2y_1=0$ which is equivalent to $z_1$ and $z_2$ are colinear
answered Jan 29 at 22:10
marwalixmarwalix
13.3k12338
$endgroup$
add a comment |
$begingroup$
Let’s assume
$$z_1=x_1+iy_1$$
$$z_2=x_2+iy_2$$
Computing one gets
$$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=x_1^2+y_1^2+x_2^2+y_2^2+2(x_1x_2+y_1y_2)$$
And
$$(|z_1|+|z_2|)^2=x_1^2+y_1^2+x_2^2+y_2^2+2|z_1z_2|$$
The identity we are investigating is therefore equivalent to
$$|z_1z_2|=(x_1x_2+y_1y_2)$$
Equivalently
$$|z_1z_2|^2=(x_1x_2+y_1y_2)^2$$
Now $z_1z_2=x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)$ which means that
$$|z_1z_2|^2=(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2$$
After expanding and identifying, one gets
$$(x_1y_2)^2+(x_2y_1)^2-2x_1x_2y_1y_2=(x_1y_2-x_2y_1)^2=0$$
And this tells us that $x_1y_2-x_2y_1=0$ which is equivalent to $z_1$ and $z_2$ are colinear
answered Jan 29 at 22:10
marwalixmarwalix
13.3k12338
$endgroup$
add a comment |
$begingroup$
Let’s assume
$$z_1=x_1+iy_1$$
$$z_2=x_2+iy_2$$
Computing one gets
$$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=x_1^2+y_1^2+x_2^2+y_2^2+2(x_1x_2+y_1y_2)$$
And
$$(|z_1|+|z_2|)^2=x_1^2+y_1^2+x_2^2+y_2^2+2|z_1z_2|$$
The identity we are investigating is therefore equivalent to
$$|z_1z_2|=(x_1x_2+y_1y_2)$$
Equivalently
$$|z_1z_2|^2=(x_1x_2+y_1y_2)^2$$
Now $z_1z_2=x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)$ which means that
$$|z_1z_2|^2=(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2$$
After expanding and identifying, one gets
$$(x_1y_2)^2+(x_2y_1)^2-2x_1x_2y_1y_2=(x_1y_2-x_2y_1)^2=0$$
And this tells us that $x_1y_2-x_2y_1=0$ which is equivalent to $z_1$ and $z_2$ are colinear
answered Jan 29 at 22:10
marwalixmarwalix
13.3k12338
$endgroup$
Let’s assume
$$z_1=x_1+iy_1$$
$$z_2=x_2+iy_2$$
Computing one gets
$$|z_1+z_2|^2=(x_1+x_2)^2+(y_1+y_2)^2=x_1^2+y_1^2+x_2^2+y_2^2+2(x_1x_2+y_1y_2)$$
And
$$(|z_1|+|z_2|)^2=x_1^2+y_1^2+x_2^2+y_2^2+2|z_1z_2|$$
The identity we are investigating is therefore equivalent to
$$|z_1z_2|=(x_1x_2+y_1y_2)$$
Equivalently
$$|z_1z_2|^2=(x_1x_2+y_1y_2)^2$$
Now $z_1z_2=x_1x_2-y_1y_2+i(x_1y_2+x_2y_1)$ which means that
$$|z_1z_2|^2=(x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2$$
After expanding and identifying, one gets
$$(x_1y_2)^2+(x_2y_1)^2-2x_1x_2y_1y_2=(x_1y_2-x_2y_1)^2=0$$
And this tells us that $x_1y_2-x_2y_1=0$ which is equivalent to $z_1$ and $z_2$ are colinear
answered Jan 29 at 22:10
marwalixmarwalix
13.3k12338
answered Jan 29 at 22:10
marwalixmarwalix
13.3k12338
answered Jan 29 at 22:10
marwalixmarwalix
13.3k12338
answered Jan 29 at 22:10
marwalixmarwalix
13.3k12338
13.3k12338
add a comment |
add a comment |
$begingroup$
I will answer the question in the heading (not including $0$).
Let $z_k=x_k+iy_k$ Square both sides and remove common terms to get: $(x_1+iy_1)(x_2-iy_2)+(x_1-iy_1)(x_2+iy_2)=(2(x_1x_2+y_1y_2))=2sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$. Now square both sides and eliminate common terms to get $2x_1x_2y_1y_2=x_1^2y_2^2+x_2^2y_1^2$ Assume none of the terms $=0$, divide by the product and get $2=frac{x_1y_2}{x_2y_1}+frac{x_2y_1}{x_1y_2}$ Notice that the two terms on the right are reciprocal, su that with $u=$ one ratio, we have a quadratic $u^2-2u+1=0$, so that the ratio $=1$, forcing the vectors to be scalar multiples of each other.
answered Jan 29 at 22:21
herb steinbergherb steinberg
3,0782311
$endgroup$
add a comment |
$begingroup$
I will answer the question in the heading (not including $0$).
Let $z_k=x_k+iy_k$ Square both sides and remove common terms to get: $(x_1+iy_1)(x_2-iy_2)+(x_1-iy_1)(x_2+iy_2)=(2(x_1x_2+y_1y_2))=2sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$. Now square both sides and eliminate common terms to get $2x_1x_2y_1y_2=x_1^2y_2^2+x_2^2y_1^2$ Assume none of the terms $=0$, divide by the product and get $2=frac{x_1y_2}{x_2y_1}+frac{x_2y_1}{x_1y_2}$ Notice that the two terms on the right are reciprocal, su that with $u=$ one ratio, we have a quadratic $u^2-2u+1=0$, so that the ratio $=1$, forcing the vectors to be scalar multiples of each other.
answered Jan 29 at 22:21
herb steinbergherb steinberg
3,0782311
$endgroup$
add a comment |
$begingroup$
I will answer the question in the heading (not including $0$).
Let $z_k=x_k+iy_k$ Square both sides and remove common terms to get: $(x_1+iy_1)(x_2-iy_2)+(x_1-iy_1)(x_2+iy_2)=(2(x_1x_2+y_1y_2))=2sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$. Now square both sides and eliminate common terms to get $2x_1x_2y_1y_2=x_1^2y_2^2+x_2^2y_1^2$ Assume none of the terms $=0$, divide by the product and get $2=frac{x_1y_2}{x_2y_1}+frac{x_2y_1}{x_1y_2}$ Notice that the two terms on the right are reciprocal, su that with $u=$ one ratio, we have a quadratic $u^2-2u+1=0$, so that the ratio $=1$, forcing the vectors to be scalar multiples of each other.
answered Jan 29 at 22:21
herb steinbergherb steinberg
3,0782311
$endgroup$
I will answer the question in the heading (not including $0$).
Let $z_k=x_k+iy_k$ Square both sides and remove common terms to get: $(x_1+iy_1)(x_2-iy_2)+(x_1-iy_1)(x_2+iy_2)=(2(x_1x_2+y_1y_2))=2sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$. Now square both sides and eliminate common terms to get $2x_1x_2y_1y_2=x_1^2y_2^2+x_2^2y_1^2$ Assume none of the terms $=0$, divide by the product and get $2=frac{x_1y_2}{x_2y_1}+frac{x_2y_1}{x_1y_2}$ Notice that the two terms on the right are reciprocal, su that with $u=$ one ratio, we have a quadratic $u^2-2u+1=0$, so that the ratio $=1$, forcing the vectors to be scalar multiples of each other.
answered Jan 29 at 22:21
herb steinbergherb steinberg
3,0782311
answered Jan 29 at 22:21
herb steinbergherb steinberg
3,0782311
answered Jan 29 at 22:21
herb steinbergherb steinberg
3,0782311
answered Jan 29 at 22:21
herb steinbergherb steinberg
3,0782311
3,0782311
add a comment |
add a comment |
$begingroup$
Let $z_1=r_1e^{itheta_1}=r_1costheta_1+ir_1sintheta_1, z_2=r_2e^{itheta_2}=r_2costheta_2+ir_2sintheta_2$.
Then $mid z_1+z_2mid=mid (r_1costheta_1 +r_2costheta_2)+i(r_1sintheta_1 +r_2sintheta_2)mid=sqrt{r_1^2cos^2theta_1 +2r_1r_2costheta_1costheta_2+r_2^2cos^2theta_2+r_1^2sin^2theta_1
+2r_1r_2sintheta_1 sintheta_2 +r_2^2sin^2theta_2}=sqrt{r_1^2+r_2^2+2r_1r_2(costheta_1, sintheta_1) cdot (costheta_2,sintheta_2) }=r_1+r_2implies (costheta_1, sintheta_1) cdot (costheta_2, sintheta_2)=1implies theta_1 =pmtheta_2 $.
Now I know you said you didn't want to use "arg", but all I used was polar coordinates. And the statement is really about the arguments of the two numbers.
answered Jan 29 at 22:27
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand.
$endgroup$
– Masie
Jan 30 at 2:48
$begingroup$
Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand.
$endgroup$
– Chris Custer
Jan 30 at 3:07
add a comment |
$begingroup$
Let $z_1=r_1e^{itheta_1}=r_1costheta_1+ir_1sintheta_1, z_2=r_2e^{itheta_2}=r_2costheta_2+ir_2sintheta_2$.
Then $mid z_1+z_2mid=mid (r_1costheta_1 +r_2costheta_2)+i(r_1sintheta_1 +r_2sintheta_2)mid=sqrt{r_1^2cos^2theta_1 +2r_1r_2costheta_1costheta_2+r_2^2cos^2theta_2+r_1^2sin^2theta_1
+2r_1r_2sintheta_1 sintheta_2 +r_2^2sin^2theta_2}=sqrt{r_1^2+r_2^2+2r_1r_2(costheta_1, sintheta_1) cdot (costheta_2,sintheta_2) }=r_1+r_2implies (costheta_1, sintheta_1) cdot (costheta_2, sintheta_2)=1implies theta_1 =pmtheta_2 $.
Now I know you said you didn't want to use "arg", but all I used was polar coordinates. And the statement is really about the arguments of the two numbers.
answered Jan 29 at 22:27
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand.
$endgroup$
– Masie
Jan 30 at 2:48
$begingroup$
Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand.
$endgroup$
– Chris Custer
Jan 30 at 3:07
add a comment |
$begingroup$
Let $z_1=r_1e^{itheta_1}=r_1costheta_1+ir_1sintheta_1, z_2=r_2e^{itheta_2}=r_2costheta_2+ir_2sintheta_2$.
Then $mid z_1+z_2mid=mid (r_1costheta_1 +r_2costheta_2)+i(r_1sintheta_1 +r_2sintheta_2)mid=sqrt{r_1^2cos^2theta_1 +2r_1r_2costheta_1costheta_2+r_2^2cos^2theta_2+r_1^2sin^2theta_1
+2r_1r_2sintheta_1 sintheta_2 +r_2^2sin^2theta_2}=sqrt{r_1^2+r_2^2+2r_1r_2(costheta_1, sintheta_1) cdot (costheta_2,sintheta_2) }=r_1+r_2implies (costheta_1, sintheta_1) cdot (costheta_2, sintheta_2)=1implies theta_1 =pmtheta_2 $.
Now I know you said you didn't want to use "arg", but all I used was polar coordinates. And the statement is really about the arguments of the two numbers.
answered Jan 29 at 22:27
Chris CusterChris Custer
14.3k3827
$endgroup$
Let $z_1=r_1e^{itheta_1}=r_1costheta_1+ir_1sintheta_1, z_2=r_2e^{itheta_2}=r_2costheta_2+ir_2sintheta_2$.
Then $mid z_1+z_2mid=mid (r_1costheta_1 +r_2costheta_2)+i(r_1sintheta_1 +r_2sintheta_2)mid=sqrt{r_1^2cos^2theta_1 +2r_1r_2costheta_1costheta_2+r_2^2cos^2theta_2+r_1^2sin^2theta_1
+2r_1r_2sintheta_1 sintheta_2 +r_2^2sin^2theta_2}=sqrt{r_1^2+r_2^2+2r_1r_2(costheta_1, sintheta_1) cdot (costheta_2,sintheta_2) }=r_1+r_2implies (costheta_1, sintheta_1) cdot (costheta_2, sintheta_2)=1implies theta_1 =pmtheta_2 $.
Now I know you said you didn't want to use "arg", but all I used was polar coordinates. And the statement is really about the arguments of the two numbers.
answered Jan 29 at 22:27
Chris CusterChris Custer
14.3k3827
edited Jan 29 at 23:20
answered Jan 29 at 22:27
Chris CusterChris Custer
14.3k3827
answered Jan 29 at 22:27
Chris CusterChris Custer
14.3k3827
answered Jan 29 at 22:27
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand.
$endgroup$
– Masie
Jan 30 at 2:48
$begingroup$
Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand.
$endgroup$
– Chris Custer
Jan 30 at 3:07
add a comment |
$begingroup$
I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand.
$endgroup$
– Masie
Jan 30 at 2:48
$begingroup$
Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand.
$endgroup$
– Chris Custer
Jan 30 at 3:07
$begingroup$
I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand.
$endgroup$
– Masie
Jan 30 at 2:48
$begingroup$
I guess I should have specified that I'm not quite comfortable using Euler's formula and exponential form as well, but this is quite clever and I do understand.
$endgroup$
– Masie
Jan 30 at 2:48
$begingroup$
Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand.
$endgroup$
– Chris Custer
Jan 30 at 3:07
$begingroup$
Great. I figured I probably shouldn't have done it. Rather, I should have tried with just $z=x+iy$. Maybe this can be done. But you will learn in short order that a complex number can be specified by an angle and a modulus, as well as the aforementioned Euler's formula, both of which are extremely useful. Glad you understand.
$endgroup$
– Chris Custer
Jan 30 at 3:07
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$begingroup$
Actually antiparallel vectors won't satisfy the equation (unless you consider 0 and any complex number to be antiparallel).
$endgroup$
– Daniel Schepler
Jan 29 at 22:03