Mixing
If $|z_1 - z_2| = |z_1 + z_2|$, then $|arg z_1 - arg z_2 |= frac{pi}{2} $ or $frac{3pi}{2}$
$begingroup$
If for $z_1, z_2in Bbb C $, $|z_1 - z_2| = |z_1 + z_2|$, then we have to prove $|arg z_1 - arg z_2| = frac{pi}{2} $ or $frac{3pi}{2}$.
I have seen similar type question here
If $|z_1 - z_2| = |z_1 + z_2|$, then $arg z_1 - arg z_2 = pi/4 $
But it is not clear to me how to show the result.
Progress :
$
|z_1 - z_2| = |z_1 + z_2| \
Rightarrow|z_1 - z_2|^2 = |z_1 + z_2|^2 \
Rightarrow (z_1-z_2)(bar z_1 - bar z_2)=(z_1+z_2)(bar z_1 + bar z_2) \
Rightarrow z_1bar z_2=-bar z_1 z_2
$
What to do now?
complex-numbers
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 3 '15 at 17:38
MKSMKS
255
$endgroup$
add a comment |
$begingroup$
If for $z_1, z_2in Bbb C $, $|z_1 - z_2| = |z_1 + z_2|$, then we have to prove $|arg z_1 - arg z_2| = frac{pi}{2} $ or $frac{3pi}{2}$.
I have seen similar type question here
If $|z_1 - z_2| = |z_1 + z_2|$, then $arg z_1 - arg z_2 = pi/4 $
But it is not clear to me how to show the result.
Progress :
$
|z_1 - z_2| = |z_1 + z_2| \
Rightarrow|z_1 - z_2|^2 = |z_1 + z_2|^2 \
Rightarrow (z_1-z_2)(bar z_1 - bar z_2)=(z_1+z_2)(bar z_1 + bar z_2) \
Rightarrow z_1bar z_2=-bar z_1 z_2
$
What to do now?
complex-numbers
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 3 '15 at 17:38
MKSMKS
255
$endgroup$
$begingroup$
It might help to let $z_{1}$ and $z_{2}$ be two points in the complex plane and draw a diagram, illustrating the addition and subtraction of the complex numbers as vectors.
$endgroup$
– Will R
Sep 3 '15 at 17:41
$begingroup$
you assume but do not state that $0 neq z_1 neq z_2 neq 0$
$endgroup$
– DanielWainfleet
Sep 3 '15 at 21:04
add a comment |
$begingroup$
If for $z_1, z_2in Bbb C $, $|z_1 - z_2| = |z_1 + z_2|$, then we have to prove $|arg z_1 - arg z_2| = frac{pi}{2} $ or $frac{3pi}{2}$.
I have seen similar type question here
If $|z_1 - z_2| = |z_1 + z_2|$, then $arg z_1 - arg z_2 = pi/4 $
But it is not clear to me how to show the result.
Progress :
$
|z_1 - z_2| = |z_1 + z_2| \
Rightarrow|z_1 - z_2|^2 = |z_1 + z_2|^2 \
Rightarrow (z_1-z_2)(bar z_1 - bar z_2)=(z_1+z_2)(bar z_1 + bar z_2) \
Rightarrow z_1bar z_2=-bar z_1 z_2
$
What to do now?
complex-numbers
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 3 '15 at 17:38
MKSMKS
255
$endgroup$
If for $z_1, z_2in Bbb C $, $|z_1 - z_2| = |z_1 + z_2|$, then we have to prove $|arg z_1 - arg z_2| = frac{pi}{2} $ or $frac{3pi}{2}$.
I have seen similar type question here
If $|z_1 - z_2| = |z_1 + z_2|$, then $arg z_1 - arg z_2 = pi/4 $
But it is not clear to me how to show the result.
Progress :
$
|z_1 - z_2| = |z_1 + z_2| \
Rightarrow|z_1 - z_2|^2 = |z_1 + z_2|^2 \
Rightarrow (z_1-z_2)(bar z_1 - bar z_2)=(z_1+z_2)(bar z_1 + bar z_2) \
Rightarrow z_1bar z_2=-bar z_1 z_2
$
What to do now?
complex-numbers
complex-numbers
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 3 '15 at 17:38
MKSMKS
255
edited Apr 13 '17 at 12:21
Community♦
1
asked Sep 3 '15 at 17:38
MKSMKS
255
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Sep 3 '15 at 17:38
MKSMKS
255
asked Sep 3 '15 at 17:38
MKSMKS
255
asked Sep 3 '15 at 17:38
MKSMKS
255
255
$begingroup$
It might help to let $z_{1}$ and $z_{2}$ be two points in the complex plane and draw a diagram, illustrating the addition and subtraction of the complex numbers as vectors.
$endgroup$
– Will R
Sep 3 '15 at 17:41
$begingroup$
you assume but do not state that $0 neq z_1 neq z_2 neq 0$
$endgroup$
– DanielWainfleet
Sep 3 '15 at 21:04
add a comment |
$begingroup$
It might help to let $z_{1}$ and $z_{2}$ be two points in the complex plane and draw a diagram, illustrating the addition and subtraction of the complex numbers as vectors.
$endgroup$
– Will R
Sep 3 '15 at 17:41
$begingroup$
you assume but do not state that $0 neq z_1 neq z_2 neq 0$
$endgroup$
– DanielWainfleet
Sep 3 '15 at 21:04
$begingroup$
It might help to let $z_{1}$ and $z_{2}$ be two points in the complex plane and draw a diagram, illustrating the addition and subtraction of the complex numbers as vectors.
$endgroup$
– Will R
Sep 3 '15 at 17:41
$begingroup$
It might help to let $z_{1}$ and $z_{2}$ be two points in the complex plane and draw a diagram, illustrating the addition and subtraction of the complex numbers as vectors.
$endgroup$
– Will R
Sep 3 '15 at 17:41
$begingroup$
you assume but do not state that $0 neq z_1 neq z_2 neq 0$
$endgroup$
– DanielWainfleet
Sep 3 '15 at 21:04
$begingroup$
you assume but do not state that $0 neq z_1 neq z_2 neq 0$
$endgroup$
– DanielWainfleet
Sep 3 '15 at 21:04
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Multiply by $z_1 z_2$ to get $z_1^2 |z_2|^2 = - z_2^2 |z_1|^2$.
Let $w_1 = z_1/|z_1|$ and $w_2 = z_2/|z_2|$ so that both $w$'s have unit length, and $arg w_i = arg z_i$.
It follows that $w_1^2 = - w_2^2$, so that $w_1 = pm i w_2$, implying $w_1$ and $w_2$ are orthogonal.
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
$endgroup$
add a comment |
$begingroup$
We may exploit the involutive property of a Cayley transform. Let $w=frac{z_2}{z_1}$. We have:
$$left|frac{1-w}{1+w}right|=1 $$
hence:
$$ frac{1-w}{1+w}=e^{itheta} $$
for some $thetainmathbb{R}$, from which $w = frac{1-e^{itheta}}{1+e^{itheta}}$ or:
$$ w = frac{e^{-itheta/2}-e^{itheta /2}}{e^{-itheta/2}+e^{itheta /2}} = -itanleft(frac{theta}{2}right).$$
That implies $arg w=pm arg i=pmfrac{pi}{2}$ and the claim follows.
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
add a comment |
$begingroup$
When $0neq z_1 neq z_2 ne 0 $ , the points $ 0, z_1, z_2 , z_1+z_2$ are the vertices of a parallelogram $P$in the complex plane. Then $|z_1-z_2| =|z_1 +z_2|$ geometrically means that the diagonals of $P$ are of equal length, which implies that $P$ is a square.
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
This question can be better viewed geometrically, but I will give you the solution from the step:
$$overline{z_{1}}z_{2}+z_{1}overline{z_{2}}=0$$
Now we know that this is nothing but :
$$2Re({z_{1}overline{z_{2}}})=0$$
So the complex number ${z_{1}overline{z_{2}}}$ is purely imaginary. When a complex number is purely imaginary the complex number is lying either on the positive or the negative imaginary axis on the Argand/Complex plane.
So the angle or the argument extended by the imaginary number ${z_{1}overline{z_{2}}}$ is either $pi/2$ or $3pi/2$.
This means that :
$$arg(z_{1}overline{z_{2}}) = pi/2 $$
or
$$arg(z_{1}overline{z_{2}}) = 3pi/2$$
Using the property,
$$arg(z_{1}overline{z_{2}}) = arg(z1)+arg(overline{z_{2}}) = arg(z_1) - arg(z_2)$$
We get :
$$ arg(z_1) - arg(z_2)= pi/2 ,or, 3pi/2$$
This same procedure has to be repeated for
$$2Re(overline{z_{1}}z_2)=0$$ and you will get the same result with a negative sign that is :
$$ arg(z_2) - arg(z_1)= pi/2 ,or, 3pi/2$$
So combining these two we get :
$$|arg(z_1) - arg(z_2)|= pi/2 ,or, 3pi/2$$
I have not considered the principal argument so $pi/2$ and $3pi/2$ will do.
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
$endgroup$
add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
votes
active
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votes
active
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votes
$begingroup$
Multiply by $z_1 z_2$ to get $z_1^2 |z_2|^2 = - z_2^2 |z_1|^2$.
Let $w_1 = z_1/|z_1|$ and $w_2 = z_2/|z_2|$ so that both $w$'s have unit length, and $arg w_i = arg z_i$.
It follows that $w_1^2 = - w_2^2$, so that $w_1 = pm i w_2$, implying $w_1$ and $w_2$ are orthogonal.
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
$endgroup$
add a comment |
$begingroup$
Multiply by $z_1 z_2$ to get $z_1^2 |z_2|^2 = - z_2^2 |z_1|^2$.
Let $w_1 = z_1/|z_1|$ and $w_2 = z_2/|z_2|$ so that both $w$'s have unit length, and $arg w_i = arg z_i$.
It follows that $w_1^2 = - w_2^2$, so that $w_1 = pm i w_2$, implying $w_1$ and $w_2$ are orthogonal.
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
$endgroup$
add a comment |
$begingroup$
Multiply by $z_1 z_2$ to get $z_1^2 |z_2|^2 = - z_2^2 |z_1|^2$.
Let $w_1 = z_1/|z_1|$ and $w_2 = z_2/|z_2|$ so that both $w$'s have unit length, and $arg w_i = arg z_i$.
It follows that $w_1^2 = - w_2^2$, so that $w_1 = pm i w_2$, implying $w_1$ and $w_2$ are orthogonal.
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
$endgroup$
Multiply by $z_1 z_2$ to get $z_1^2 |z_2|^2 = - z_2^2 |z_1|^2$.
Let $w_1 = z_1/|z_1|$ and $w_2 = z_2/|z_2|$ so that both $w$'s have unit length, and $arg w_i = arg z_i$.
It follows that $w_1^2 = - w_2^2$, so that $w_1 = pm i w_2$, implying $w_1$ and $w_2$ are orthogonal.
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
answered Sep 3 '15 at 17:45
Umberto P.Umberto P.
39.6k13267
39.6k13267
add a comment |
add a comment |
$begingroup$
We may exploit the involutive property of a Cayley transform. Let $w=frac{z_2}{z_1}$. We have:
$$left|frac{1-w}{1+w}right|=1 $$
hence:
$$ frac{1-w}{1+w}=e^{itheta} $$
for some $thetainmathbb{R}$, from which $w = frac{1-e^{itheta}}{1+e^{itheta}}$ or:
$$ w = frac{e^{-itheta/2}-e^{itheta /2}}{e^{-itheta/2}+e^{itheta /2}} = -itanleft(frac{theta}{2}right).$$
That implies $arg w=pm arg i=pmfrac{pi}{2}$ and the claim follows.
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
add a comment |
$begingroup$
We may exploit the involutive property of a Cayley transform. Let $w=frac{z_2}{z_1}$. We have:
$$left|frac{1-w}{1+w}right|=1 $$
hence:
$$ frac{1-w}{1+w}=e^{itheta} $$
for some $thetainmathbb{R}$, from which $w = frac{1-e^{itheta}}{1+e^{itheta}}$ or:
$$ w = frac{e^{-itheta/2}-e^{itheta /2}}{e^{-itheta/2}+e^{itheta /2}} = -itanleft(frac{theta}{2}right).$$
That implies $arg w=pm arg i=pmfrac{pi}{2}$ and the claim follows.
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
add a comment |
$begingroup$
We may exploit the involutive property of a Cayley transform. Let $w=frac{z_2}{z_1}$. We have:
$$left|frac{1-w}{1+w}right|=1 $$
hence:
$$ frac{1-w}{1+w}=e^{itheta} $$
for some $thetainmathbb{R}$, from which $w = frac{1-e^{itheta}}{1+e^{itheta}}$ or:
$$ w = frac{e^{-itheta/2}-e^{itheta /2}}{e^{-itheta/2}+e^{itheta /2}} = -itanleft(frac{theta}{2}right).$$
That implies $arg w=pm arg i=pmfrac{pi}{2}$ and the claim follows.
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
$endgroup$
We may exploit the involutive property of a Cayley transform. Let $w=frac{z_2}{z_1}$. We have:
$$left|frac{1-w}{1+w}right|=1 $$
hence:
$$ frac{1-w}{1+w}=e^{itheta} $$
for some $thetainmathbb{R}$, from which $w = frac{1-e^{itheta}}{1+e^{itheta}}$ or:
$$ w = frac{e^{-itheta/2}-e^{itheta /2}}{e^{-itheta/2}+e^{itheta /2}} = -itanleft(frac{theta}{2}right).$$
That implies $arg w=pm arg i=pmfrac{pi}{2}$ and the claim follows.
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
answered Sep 3 '15 at 17:54
Jack D'AurizioJack D'Aurizio
290k33284666
290k33284666
add a comment |
add a comment |
$begingroup$
When $0neq z_1 neq z_2 ne 0 $ , the points $ 0, z_1, z_2 , z_1+z_2$ are the vertices of a parallelogram $P$in the complex plane. Then $|z_1-z_2| =|z_1 +z_2|$ geometrically means that the diagonals of $P$ are of equal length, which implies that $P$ is a square.
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
When $0neq z_1 neq z_2 ne 0 $ , the points $ 0, z_1, z_2 , z_1+z_2$ are the vertices of a parallelogram $P$in the complex plane. Then $|z_1-z_2| =|z_1 +z_2|$ geometrically means that the diagonals of $P$ are of equal length, which implies that $P$ is a square.
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
When $0neq z_1 neq z_2 ne 0 $ , the points $ 0, z_1, z_2 , z_1+z_2$ are the vertices of a parallelogram $P$in the complex plane. Then $|z_1-z_2| =|z_1 +z_2|$ geometrically means that the diagonals of $P$ are of equal length, which implies that $P$ is a square.
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
When $0neq z_1 neq z_2 ne 0 $ , the points $ 0, z_1, z_2 , z_1+z_2$ are the vertices of a parallelogram $P$in the complex plane. Then $|z_1-z_2| =|z_1 +z_2|$ geometrically means that the diagonals of $P$ are of equal length, which implies that $P$ is a square.
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
answered Sep 3 '15 at 21:05
DanielWainfleetDanielWainfleet
35.3k31648
35.3k31648
add a comment |
add a comment |
$begingroup$
This question can be better viewed geometrically, but I will give you the solution from the step:
$$overline{z_{1}}z_{2}+z_{1}overline{z_{2}}=0$$
Now we know that this is nothing but :
$$2Re({z_{1}overline{z_{2}}})=0$$
So the complex number ${z_{1}overline{z_{2}}}$ is purely imaginary. When a complex number is purely imaginary the complex number is lying either on the positive or the negative imaginary axis on the Argand/Complex plane.
So the angle or the argument extended by the imaginary number ${z_{1}overline{z_{2}}}$ is either $pi/2$ or $3pi/2$.
This means that :
$$arg(z_{1}overline{z_{2}}) = pi/2 $$
or
$$arg(z_{1}overline{z_{2}}) = 3pi/2$$
Using the property,
$$arg(z_{1}overline{z_{2}}) = arg(z1)+arg(overline{z_{2}}) = arg(z_1) - arg(z_2)$$
We get :
$$ arg(z_1) - arg(z_2)= pi/2 ,or, 3pi/2$$
This same procedure has to be repeated for
$$2Re(overline{z_{1}}z_2)=0$$ and you will get the same result with a negative sign that is :
$$ arg(z_2) - arg(z_1)= pi/2 ,or, 3pi/2$$
So combining these two we get :
$$|arg(z_1) - arg(z_2)|= pi/2 ,or, 3pi/2$$
I have not considered the principal argument so $pi/2$ and $3pi/2$ will do.
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
$endgroup$
add a comment |
$begingroup$
This question can be better viewed geometrically, but I will give you the solution from the step:
$$overline{z_{1}}z_{2}+z_{1}overline{z_{2}}=0$$
Now we know that this is nothing but :
$$2Re({z_{1}overline{z_{2}}})=0$$
So the complex number ${z_{1}overline{z_{2}}}$ is purely imaginary. When a complex number is purely imaginary the complex number is lying either on the positive or the negative imaginary axis on the Argand/Complex plane.
So the angle or the argument extended by the imaginary number ${z_{1}overline{z_{2}}}$ is either $pi/2$ or $3pi/2$.
This means that :
$$arg(z_{1}overline{z_{2}}) = pi/2 $$
or
$$arg(z_{1}overline{z_{2}}) = 3pi/2$$
Using the property,
$$arg(z_{1}overline{z_{2}}) = arg(z1)+arg(overline{z_{2}}) = arg(z_1) - arg(z_2)$$
We get :
$$ arg(z_1) - arg(z_2)= pi/2 ,or, 3pi/2$$
This same procedure has to be repeated for
$$2Re(overline{z_{1}}z_2)=0$$ and you will get the same result with a negative sign that is :
$$ arg(z_2) - arg(z_1)= pi/2 ,or, 3pi/2$$
So combining these two we get :
$$|arg(z_1) - arg(z_2)|= pi/2 ,or, 3pi/2$$
I have not considered the principal argument so $pi/2$ and $3pi/2$ will do.
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
$endgroup$
add a comment |
$begingroup$
This question can be better viewed geometrically, but I will give you the solution from the step:
$$overline{z_{1}}z_{2}+z_{1}overline{z_{2}}=0$$
Now we know that this is nothing but :
$$2Re({z_{1}overline{z_{2}}})=0$$
So the complex number ${z_{1}overline{z_{2}}}$ is purely imaginary. When a complex number is purely imaginary the complex number is lying either on the positive or the negative imaginary axis on the Argand/Complex plane.
So the angle or the argument extended by the imaginary number ${z_{1}overline{z_{2}}}$ is either $pi/2$ or $3pi/2$.
This means that :
$$arg(z_{1}overline{z_{2}}) = pi/2 $$
or
$$arg(z_{1}overline{z_{2}}) = 3pi/2$$
Using the property,
$$arg(z_{1}overline{z_{2}}) = arg(z1)+arg(overline{z_{2}}) = arg(z_1) - arg(z_2)$$
We get :
$$ arg(z_1) - arg(z_2)= pi/2 ,or, 3pi/2$$
This same procedure has to be repeated for
$$2Re(overline{z_{1}}z_2)=0$$ and you will get the same result with a negative sign that is :
$$ arg(z_2) - arg(z_1)= pi/2 ,or, 3pi/2$$
So combining these two we get :
$$|arg(z_1) - arg(z_2)|= pi/2 ,or, 3pi/2$$
I have not considered the principal argument so $pi/2$ and $3pi/2$ will do.
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
$endgroup$
This question can be better viewed geometrically, but I will give you the solution from the step:
$$overline{z_{1}}z_{2}+z_{1}overline{z_{2}}=0$$
Now we know that this is nothing but :
$$2Re({z_{1}overline{z_{2}}})=0$$
So the complex number ${z_{1}overline{z_{2}}}$ is purely imaginary. When a complex number is purely imaginary the complex number is lying either on the positive or the negative imaginary axis on the Argand/Complex plane.
So the angle or the argument extended by the imaginary number ${z_{1}overline{z_{2}}}$ is either $pi/2$ or $3pi/2$.
This means that :
$$arg(z_{1}overline{z_{2}}) = pi/2 $$
or
$$arg(z_{1}overline{z_{2}}) = 3pi/2$$
Using the property,
$$arg(z_{1}overline{z_{2}}) = arg(z1)+arg(overline{z_{2}}) = arg(z_1) - arg(z_2)$$
We get :
$$ arg(z_1) - arg(z_2)= pi/2 ,or, 3pi/2$$
This same procedure has to be repeated for
$$2Re(overline{z_{1}}z_2)=0$$ and you will get the same result with a negative sign that is :
$$ arg(z_2) - arg(z_1)= pi/2 ,or, 3pi/2$$
So combining these two we get :
$$|arg(z_1) - arg(z_2)|= pi/2 ,or, 3pi/2$$
I have not considered the principal argument so $pi/2$ and $3pi/2$ will do.
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
answered Jan 20 at 19:17
SNEHIL SANYALSNEHIL SANYAL
610110
610110
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$begingroup$
It might help to let $z_{1}$ and $z_{2}$ be two points in the complex plane and draw a diagram, illustrating the addition and subtraction of the complex numbers as vectors.
$endgroup$
– Will R
Sep 3 '15 at 17:41
$begingroup$
you assume but do not state that $0 neq z_1 neq z_2 neq 0$
$endgroup$
– DanielWainfleet
Sep 3 '15 at 21:04