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.
Best Answer
$z=\frac{-3+6i}{10}$ 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 $z=a+bi \implies \bar{z}=a-bi\implies \bar{z}i=b+ai$
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 $z-\bar{z}i=3\implies a-b-(a-b)i=3$ implies that the $(a-b)$ is simultaneously $3$ and $0$.
So we conclude: No solutions.