Cfrgtkky

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.

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.

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

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.

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

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


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