Find all complex numbers z such that $\Im(-z+i) = (z+i)^2$

Question

$$\Im(-z+i) = (z+i)^2$$

So I tried solving this by replacing $z=a+bi$, thus getting $$\Im(-a+(1-b)i) = (a+(b+1)i)^2$$

From here I split it into the imaginery and real part:

\begin{equation*} \begin{aligned} -a &= a^2 -(b+1)^2 \\ 1-b &= 2ab + 2a - 1 \end{aligned} \end{equation*}

Now, I can't seem to solve this set without getting a 4th degree $b$. So I'm thinking my logic elsewhere is flawed. Can anyone tell me what I'm doing wrong?

Answer 1

EDITED!

The imaginary part of a complex number is always a real number. Hence, in order to solve your equation you have to impose that

$\Im\left((z+1)^2\right)=0$

$\Re\left((z+1)^2\right)=\Im(-z+i)$

I.e. solve the following system:

$2ab+2b=2b(a+1)=0$ from which you have $b=0$ or $a=-1$.

$(a+1)^2-b^2=1-b$ that becomes: $\;\;a=0,-2$ if we choose $b=0$ in the previous line... or becomes $b^2-b+1=0$ if you choose $a=-1$ in the previous line which has no solution.

Hence you have two solutions: $z_1=0$, $z_2=-2$.

Answer 2

Let $z=a+ib$ where $a,b$ are real

$$\implies z+1=(a+1)+ib\implies (z+1)^2=(a+1)^2-b^2+2ib(a+1)$$

and $\displaystyle\Im(-a+(1-b)i) =1-b $

Now equate the imaginary part to get $b\cdot(a+1)=0$

Can you take it from here?

Answer 3

The LHS notation is saying that the imaginary component of $-z+i$ equals the RHS. You are instead solving $-z+i = (z+i)^2$, which is easier to do by solving the corresponding quadratic in $z$.

As you have written the question, the first thing to realize is that $(z+i)^2$ must be real, with no imaginary component. What restriction does this place on $z$?