Mixing
Solving complex equation: $(z-1)^2+(bar{z}-2i)^2 = 0$
up vote
0
down vote
favorite
We're supposed to solve this complex numbers equation:
$(z-1)^2+(bar{z}-2i)^2 = 0$
I'm getting the result:
$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$
Others are getting the same result. However, the answers page says that the result should be:
$z_{0} = - frac{3}{10} + frac{3}{5}i$
Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.
complex-numbers
edited 2 days ago
amWhy
191k27223437
asked 2 days ago
weno
203
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
We're supposed to solve this complex numbers equation:
$(z-1)^2+(bar{z}-2i)^2 = 0$
I'm getting the result:
$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$
Others are getting the same result. However, the answers page says that the result should be:
$z_{0} = - frac{3}{10} + frac{3}{5}i$
Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.
complex-numbers
edited 2 days ago
amWhy
191k27223437
asked 2 days ago
weno
203
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
2 days ago
1
You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We're supposed to solve this complex numbers equation:
$(z-1)^2+(bar{z}-2i)^2 = 0$
I'm getting the result:
$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$
Others are getting the same result. However, the answers page says that the result should be:
$z_{0} = - frac{3}{10} + frac{3}{5}i$
Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.
complex-numbers
edited 2 days ago
amWhy
191k27223437
asked 2 days ago
weno
203
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We're supposed to solve this complex numbers equation:
$(z-1)^2+(bar{z}-2i)^2 = 0$
I'm getting the result:
$z_{1} = frac{1-i}{2}, z_{2} = frac{1+i}{2}$
Others are getting the same result. However, the answers page says that the result should be:
$z_{0} = - frac{3}{10} + frac{3}{5}i$
Would anyone be able to give it a look and verify whether I am wrong or the answers page is wrong? Thanks.
complex-numbers
complex-numbers
edited 2 days ago
amWhy
191k27223437
asked 2 days ago
weno
203
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
amWhy
191k27223437
asked 2 days ago
weno
203
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
amWhy
191k27223437
edited 2 days ago
amWhy
191k27223437
edited 2 days ago
amWhy
191k27223437
191k27223437
asked 2 days ago
weno
203
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
weno
203
asked 2 days ago
weno
203
203
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
2 days ago
1
You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
2 days ago
add a comment |
4
Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
2 days ago
1
You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
2 days ago
4
4
Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
2 days ago
Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
2 days ago
1
1
You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
2 days ago
You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
up vote
0
down vote
accepted
There is no solution for the complex equation below is the proof:
$(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
$$2x^2-2y^2-2x-4y-3=0$$
and
$$-2y-4x=0 Rightarrow y=-2x$$
Substituting $y=-2x$ into the first real equation we get
$$2x^2-8x^2-2x+8x-3=0$$
or$$-6x^2+6x-3=0$$
or $$x^2-x+0.5=0$$
so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.
answered 2 days ago
Arash Rashidi
558
add a comment |
up vote
1
down vote
$z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.
$z-1=frac{-13+6i}{10}$,
$bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$
$$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$
$$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$
So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$
In fact, there are not solutions to this equation.
It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$
$a+bi=z-1+bar{z}+2=z+bar{z}+1$ and
$a-bi=z-1-bar{z}-2=z-bar{z}-3$ and
We arrive at
$(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$
Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$
So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
What's note worthy here is that the real and the imaginary parts of these numbers are the same.
Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.
So we conclude: No solutions.
answered 2 days ago
Mason
1,6521325
add a comment |
up vote
0
down vote
Alternative way to arrive to the conclusion that the equation has no solutions:
Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is
$$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$
$$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$
and compare now real and imaginary parts:
$$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$
$$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$
adn the last equation has no real solution...as it should if there was a solution.
answered 2 days ago
DonAntonio
175k1491224
add a comment |
up vote
0
down vote
Late answer but I think worth mentioning it:
You can write the expression as difference of squares and factor it into two factors which can never be $0$:
$$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
& = & (z+1 + ibar z)(z-3 - ibar z) \
& = & 0
end{eqnarray*}$$
With $z = a+ib$ you get
- $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
- $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
answered yesterday
trancelocation
8,1291519
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There is no solution for the complex equation below is the proof:
$(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
$$2x^2-2y^2-2x-4y-3=0$$
and
$$-2y-4x=0 Rightarrow y=-2x$$
Substituting $y=-2x$ into the first real equation we get
$$2x^2-8x^2-2x+8x-3=0$$
or$$-6x^2+6x-3=0$$
or $$x^2-x+0.5=0$$
so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.
answered 2 days ago
Arash Rashidi
558
add a comment |
up vote
0
down vote
accepted
There is no solution for the complex equation below is the proof:
$(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
$$2x^2-2y^2-2x-4y-3=0$$
and
$$-2y-4x=0 Rightarrow y=-2x$$
Substituting $y=-2x$ into the first real equation we get
$$2x^2-8x^2-2x+8x-3=0$$
or$$-6x^2+6x-3=0$$
or $$x^2-x+0.5=0$$
so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.
answered 2 days ago
Arash Rashidi
558
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There is no solution for the complex equation below is the proof:
$(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
$$2x^2-2y^2-2x-4y-3=0$$
and
$$-2y-4x=0 Rightarrow y=-2x$$
Substituting $y=-2x$ into the first real equation we get
$$2x^2-8x^2-2x+8x-3=0$$
or$$-6x^2+6x-3=0$$
or $$x^2-x+0.5=0$$
so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.
answered 2 days ago
Arash Rashidi
558
There is no solution for the complex equation below is the proof:
$(z-1)^2+(bar{z}-2i)^2 = 0 Rightarrow z^2+1-2z+bar{z^2}-4-4ibar{z}=0$
We know $z^2+bar{z^2}=2 Re{z^2}=2x^2-2y^2$ where $z=x+iy$ and $x$ and $y$ are real numbers. Substituting $z=x+iy$ in the complex equation we get $2x^2-2y^2-3-2(x+iy)-4i(x-iy)=0$. So both real and imaginary parts of the resultant equation must be zero. So we get:
$$2x^2-2y^2-2x-4y-3=0$$
and
$$-2y-4x=0 Rightarrow y=-2x$$
Substituting $y=-2x$ into the first real equation we get
$$2x^2-8x^2-2x+8x-3=0$$
or$$-6x^2+6x-3=0$$
or $$x^2-x+0.5=0$$
so $$x=frac{1pm i}{2}$$ which is not a real number. So there will not be any real $x$ and $y$ satisfying the complex equation. Therefore no complex $z$ satisfies the equation.
answered 2 days ago
Arash Rashidi
558
answered 2 days ago
Arash Rashidi
558
answered 2 days ago
Arash Rashidi
558
answered 2 days ago
Arash Rashidi
558
558
add a comment |
add a comment |
up vote
1
down vote
$z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.
$z-1=frac{-13+6i}{10}$,
$bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$
$$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$
$$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$
So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$
In fact, there are not solutions to this equation.
It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$
$a+bi=z-1+bar{z}+2=z+bar{z}+1$ and
$a-bi=z-1-bar{z}-2=z-bar{z}-3$ and
We arrive at
$(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$
Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$
So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
What's note worthy here is that the real and the imaginary parts of these numbers are the same.
Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.
So we conclude: No solutions.
answered 2 days ago
Mason
1,6521325
add a comment |
up vote
1
down vote
$z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.
$z-1=frac{-13+6i}{10}$,
$bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$
$$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$
$$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$
So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$
In fact, there are not solutions to this equation.
It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$
$a+bi=z-1+bar{z}+2=z+bar{z}+1$ and
$a-bi=z-1-bar{z}-2=z-bar{z}-3$ and
We arrive at
$(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$
Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$
So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
What's note worthy here is that the real and the imaginary parts of these numbers are the same.
Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.
So we conclude: No solutions.
answered 2 days ago
Mason
1,6521325
add a comment |
up vote
1
down vote
up vote
1
down vote
$z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.
$z-1=frac{-13+6i}{10}$,
$bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$
$$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$
$$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$
So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$
In fact, there are not solutions to this equation.
It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$
$a+bi=z-1+bar{z}+2=z+bar{z}+1$ and
$a-bi=z-1-bar{z}-2=z-bar{z}-3$ and
We arrive at
$(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$
Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$
So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
What's note worthy here is that the real and the imaginary parts of these numbers are the same.
Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.
So we conclude: No solutions.
answered 2 days ago
Mason
1,6521325
$z=frac{-3+6i}{10}$ is not a solution to this problem. Let's make sure.
$z-1=frac{-13+6i}{10}$,
$bar{z}-2i=frac{-3-6i}{10}-frac{20i}{10} =frac{-3-26i}{10}$
$$(z-1)^2=frac{(-13+6i)^2}{100}=frac{169 - 156i-36}{100}=frac{133-156i}{100}$$
$$(bar{z}-2i)^2 =frac{(-3-26i)^2}{100}=frac{9-156i-676}{100}=frac{156i-667}{100}$$
So therefore $(z-1)^2+(bar{z}-2i)^2=-5.34$
In fact, there are not solutions to this equation.
It's not that hard to see if you apply $a^2+b^2=(a+bi)(a-bi)$
Taking $a=z-1$ and $b=bar{z}-2i implies bi= bar{z}i+2$
$a+bi=z-1+bar{z}+2=z+bar{z}+1$ and
$a-bi=z-1-bar{z}-2=z-bar{z}-3$ and
We arrive at
$(z-1)^2+(bar{z}-2i)^2=(z+bar{z}i+1)(z-bar{z}i-3)=0$
Taking $z=a+bi implies bar{z}=a-biimplies bar{z}i=b+ai$
So then $z+bar{z}i=(a+b)+(a+b)i$ and $z-bar{z}i= a-b-(a-b)i$
What's note worthy here is that the real and the imaginary parts of these numbers are the same.
Then $z+bar{z}i=-1 implies a+b+(a+b)i=-1$ but then $a+b$ is simulataneously $0$ and $-1$ and likewise $z-bar{z}i=3implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.
So we conclude: No solutions.
answered 2 days ago
Mason
1,6521325
edited 2 days ago
answered 2 days ago
Mason
1,6521325
answered 2 days ago
Mason
1,6521325
answered 2 days ago
Mason
1,6521325
1,6521325
add a comment |
add a comment |
up vote
0
down vote
Alternative way to arrive to the conclusion that the equation has no solutions:
Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is
$$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$
$$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$
and compare now real and imaginary parts:
$$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$
$$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$
adn the last equation has no real solution...as it should if there was a solution.
answered 2 days ago
DonAntonio
175k1491224
add a comment |
up vote
0
down vote
Alternative way to arrive to the conclusion that the equation has no solutions:
Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is
$$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$
$$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$
and compare now real and imaginary parts:
$$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$
$$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$
adn the last equation has no real solution...as it should if there was a solution.
answered 2 days ago
DonAntonio
175k1491224
add a comment |
up vote
0
down vote
up vote
0
down vote
Alternative way to arrive to the conclusion that the equation has no solutions:
Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is
$$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$
$$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$
and compare now real and imaginary parts:
$$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$
$$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$
adn the last equation has no real solution...as it should if there was a solution.
answered 2 days ago
DonAntonio
175k1491224
Alternative way to arrive to the conclusion that the equation has no solutions:
Write $;z=a+biimpliesoverline z=a-bi;$ , and the given equation is
$$(a-1+bi)^2+(a-(b+2)i)^2=0iff$$
$$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$
and compare now real and imaginary parts:
$$begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0iff2a^2-2b^2-2a-4b-3=0\
2ab-2b-2ab-4a=0iff b=-2aend{cases};;;implies$$
$$2a^2-8a^2-2a+8a-3=0iff 6a^2-6a+3=0$$
adn the last equation has no real solution...as it should if there was a solution.
answered 2 days ago
DonAntonio
175k1491224
answered 2 days ago
DonAntonio
175k1491224
answered 2 days ago
DonAntonio
175k1491224
answered 2 days ago
DonAntonio
175k1491224
175k1491224
add a comment |
add a comment |
up vote
0
down vote
Late answer but I think worth mentioning it:
You can write the expression as difference of squares and factor it into two factors which can never be $0$:
$$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
& = & (z+1 + ibar z)(z-3 - ibar z) \
& = & 0
end{eqnarray*}$$
With $z = a+ib$ you get
- $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
- $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
answered yesterday
trancelocation
8,1291519
add a comment |
up vote
0
down vote
Late answer but I think worth mentioning it:
You can write the expression as difference of squares and factor it into two factors which can never be $0$:
$$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
& = & (z+1 + ibar z)(z-3 - ibar z) \
& = & 0
end{eqnarray*}$$
With $z = a+ib$ you get
- $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
- $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
answered yesterday
trancelocation
8,1291519
add a comment |
up vote
0
down vote
up vote
0
down vote
Late answer but I think worth mentioning it:
You can write the expression as difference of squares and factor it into two factors which can never be $0$:
$$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
& = & (z+1 + ibar z)(z-3 - ibar z) \
& = & 0
end{eqnarray*}$$
With $z = a+ib$ you get
- $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
- $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
answered yesterday
trancelocation
8,1291519
Late answer but I think worth mentioning it:
You can write the expression as difference of squares and factor it into two factors which can never be $0$:
$$begin{eqnarray*}(z-1)^2+(bar{z}-2i)^2 & = & (z-1)^2-i^2(bar{z}-2i)^2 \
& = & (z+1 + ibar z)(z-3 - ibar z) \
& = & 0
end{eqnarray*}$$
With $z = a+ib$ you get
- $z+1 + ibar z = a+b+1 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
- $z-3 - ibar z =a+b-3 + i(a+b) stackrel{!}{=} 0 Rightarrow mbox{No solution!}$
answered yesterday
trancelocation
8,1291519
answered yesterday
trancelocation
8,1291519
answered yesterday
trancelocation
8,1291519
answered yesterday
trancelocation
8,1291519
8,1291519
add a comment |
add a comment |
weno is a new contributor. Be nice, and check out our Code of Conduct.
weno is a new contributor. Be nice, and check out our Code of Conduct.
weno is a new contributor. Be nice, and check out our Code of Conduct.
weno is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005312%2fsolving-complex-equation-z-12-barz-2i2-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
Please add the work you did to get the results you report that you obtained. Expand the equation, and show us your work in determining $z_1, z_2$. Whether the results are correct or not isn't as important, and can't be substituted, for how you obtained them. Furthermore, adding your work, in the event that you are wrong, will help us help you identify where your mistake may have been.
– amWhy
2 days ago
1
You can plug in your numbers to show that $(1-i)/2$ and $(1+i)/2$ do not satisfy the original equation, and hence are not the solutions. I would probably set $z=a+bi,$ with $a$ and $b$ real, and solve the two equations simultaneously.
– Adrian Keister
2 days ago