Find the real and imaginary parts of (1-z)/(i+z)

[carlosbgois]

Member advised that the homework template is required.
Hey there! Need help figuring this out:
Find the real and imaginary parts of \frac{1-z}{i+z}

What I've tried was to notice that z\bar{z}=|z|^2, thence \frac{1-z}{i+z}=\frac{(1-z)}{(i+z)}\frac{(\overline{i+z})}{(\overline{i+z})}=\frac{(1-z)(i+\overline{z})}{|i+z|^2}=\frac{\overline{z}+i(z-1)-|z|^2}{|i+z|^2}

But now I'm stuck. Any help is appreciated. Thanks in advance.

 

[PeroK]

What sort of answer are you looking for? Usually you would have $z = x + iy$ and would express the real and imaginary parts of the expression in terms of $x$ and $y$.

 

[Mark44]

Hey there! Need help figuring this out:
Find the real and imaginary parts of \frac{1-z}{i+z}

Let z = x + iy, and write the above as $\frac{1 - x - iy}{i + x + iy} = \frac{1 - x - iy}{x + i(y + 1)}$, and then multiply by 1 in the form of the conjugate over itself.

Also, in future posts, please don't delete the homework template - it's not optional.

What I've tried was to notice that z\bar{z}=|z|^2, thence \frac{1-z}{i+z}=\frac{(1-z)}{(i+z)}\frac{(\overline{i+z})}{(\overline{i+z})}=\frac{(1-z)(i+\overline{z})}{|i+z|^2}=\frac{\overline{z}+i(z-1)-|z|^2}{|i+z|^2}

But now I'm stuck. Any help is appreciated. Thanks in advance.