$\text{Im}(z)$ in equation

Question

I'm having trouble with this equation:

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

After a bit of algebra i've gotten:

$$1-\text{Im}(z) = z^2 + 2iz - 1$$

But i have no clue where to go from here, how do i get rid of the "$\text{Im}$"?

Answer 1

Hint

Write $z=a+i~b$ in which $a$ and $b$ are real numbers. So $$Im(-z+i)=Im(-a+i(1-b))=1-b$$

Since John's answer came while I was typing, just continue the way he suggests (this is what I was about to write).

Continuation of my answer

The right hand side is $$(z+i)^2=(a+i(b+1))^2=a^2-(b+1)^2+2a(b+1)i$$ So the equation is $$1-b=a^2-(b+1)^2+2a(b+1)i$$ Now, the method consists in the identification of real and imaginary parts. This means that we have tox equations $$1-b=a^2+(b+1)^2$$ $$2a(b+1)=0$$ that is to say two equations for two unknowns $a$ and $b$.

I am sure that you can take from here.

Answer 2

Try letting $z=a+b i$ with $a,b\in \mathbb{R}$. After some simplification, you should be able to equate the real and imaginary components of each side in order to determine $a$ and $b$.