2
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ $+1$. 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? $\endgroup$ Feb 1, 2017 at 5:20
  • 1
    $\begingroup$ The key to this calculation is that $z$ and its conjugate $\overline{z}$ share the same real part. $\endgroup$
    – hardmath
    Feb 1, 2017 at 5:21
  • $\begingroup$ 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 $\Im$. $\endgroup$
    – hardmath
    Feb 1, 2017 at 5:26

2 Answers 2

2
$\begingroup$

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 $$ \frac{1}{a+ib}=\frac{1}{a+ib}\frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}=\frac{a}{a^2+b^2}-i\frac{b}{a^2+b^2}.$$

$\endgroup$
1
  • $\begingroup$ Thank you! I see where I was stuck! $\endgroup$
    – Charlotte
    Feb 1, 2017 at 5:47
1
$\begingroup$

$$Re(\frac{1}{z})=Re(\frac{1}{z}\times\frac{\bar{z}}{\bar{z}})=Re(\frac{\bar{z}}{|z|^2})=\frac{1}{|z|^2}Re(\bar{z})=\frac{1}{|z|^2}Re(z)=Re(\frac{z}{|z|^2})$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .