Solving complex equation: $(z-1)^2+(\bar{z}-2i)^2 = 0$

Question

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.

Answer 1

There is no solution for the complex equation below is the proof: $(z-1)^2+(\bar{z}-2i)^2 = 0 \Rightarrow z^2+1-2z+\bar{z^2}-4-4i\bar{z}=0$ 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 $$x=\frac{1\pm i}{2}$$ 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.

Answer 2

$z=\frac{-3+6i}{10}$ 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.

Answer 3

Alternative way to arrive to the conclusion that the equation has no solutions:

Write $\;z=a+bi\implies\overline z=a-bi\;$ , and the given equation is

$$(a-1+bi)^2+(a-(b+2)i)^2=0\iff$$

$$(a-1)^2-b^2+2(a-1)bi+a^2-(b+2)^2-2a(b+2)i=0$$

and compare now real and imaginary parts:

$$\begin{cases}(a-1)^2-b^2+a^2-(b+2)^2=0\iff2a^2-2b^2-2a-4b-3=0\\ 2ab-2b-2ab-4a=0\iff b=-2a\end{cases}\;\;\;\implies$$

$$2a^2-8a^2-2a+8a-3=0\iff 6a^2-6a+3=0$$

adn the last equation has no real solution...as it should if there was a solution.

Answer 4

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 $0$: $$\begin{eqnarray*}(z-1)^2+(\bar{z}-2i)^2 & = & (z-1)^2-i^2(\bar{z}-2i)^2 \\ & = & (z+1 + i\bar z)(z-3 - i\bar z) \\ & = & 0 \end{eqnarray*}$$ With $z = a+ib$ you get

• $z+1 + i\bar z = a+b+1 + i(a+b) \stackrel{!}{=} 0 \Rightarrow \mbox{No solution!}$
• $z-3 - i\bar z =a+b-3 + i(a+b) \stackrel{!}{=} 0 \Rightarrow \mbox{No solution!}$