# Basic Question on the real and imaginary part of a complex number

I was introduced to the use of Re and Im to denote the real and imaginary parts of complex numbers today, but I'm still feeling unclear on how exactly they work.

I think understand them for a basic example: $$Re(a+ib)=a$$

But we were given this equality, which I'm unsure about: $$Re\left(\frac{1}{z}\right)=Re\left(\frac{z}{{\lvert z\rvert}^2}\right)$$ If someone could help me out, maybe with a simple proof of this, it would be helpful!

• $+1$. Since you are a new user, I am happy that you have provided context to your question. Where do you think you can start with this question? Since you are only unsure, you must not be clueless, right? Feb 1, 2017 at 5:20
• The key to this calculation is that $z$ and its conjugate $\overline{z}$ share the same real part. Feb 1, 2017 at 5:21
• Note that $\LaTeX$ has fancy symbols for the real and imaginary parts. If we write \Re and \Im inside the MathJax markdown, we get: $\Re$ and $\Im$. Feb 1, 2017 at 5:26

Recall $|z|^2=z\overline{z}$ where $\overline{z}$ is the complex conjugate of $z$ (that is, if $z=a+ib$. then $\overline{z}=a-ib$).
Then use $$\frac{1}{a+ib}=\frac{1}{a+ib}\frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}=\frac{a}{a^2+b^2}-i\frac{b}{a^2+b^2}.$$
$$Re(\frac{1}{z})=Re(\frac{1}{z}\times\frac{\bar{z}}{\bar{z}})=Re(\frac{\bar{z}}{|z|^2})=\frac{1}{|z|^2}Re(\bar{z})=\frac{1}{|z|^2}Re(z)=Re(\frac{z}{|z|^2})$$