Complex conjugation

Question

Prove $\operatorname{Im}(z)=-\frac{i}{2}(z-z^\ast)$

Hey guys,

Love the site. Learning more here than in my lectures.

I’m assuming I have to let $a+bi =z$ and get $\operatorname{Im}(z)=b$ but I’m still not sure how to get this.

Any help would be much appreciated.

Answer 1

Let $z=a+ib$, then you have $z-\overline{z}=a+ib-a+ib=2ib$. From here, we have: $$-\frac{i}{2}(z-\overline{z})=-\frac{i}{2}\cdot 2ib=-i^2b=b=\Im(z)$$

Answer 2

$Im(z) = \frac{z-\bar{z}}{2i} = \frac{(z-\bar{z})i}{-2} = \frac{-i(z-\bar{z})}{2} \\$

This is enough to answer your question, but if you do want to prove that $Im(z) = \frac{z-\bar{z}}{2i},$ then let z = a+bi:

$\frac{(a+bi)-(a-bi)}{2i}=\frac{2bi}{2i}=b=Im(z)$