2022-01-17

Is $|\frac{a-ib}{a+ib}|$ equal to 1? If yes, how?

nick1337

Expert

Step 1
$|z|=|\overline{z}|$
since the vector $\overline{z}$ is a mirror copy of z , upon reflection into the real axis, which leaves its norm (length) unaltered.
The above assumes that $a,b\in \mathbb{R}$. In fact the same is not true is we have $a,b\in \mathbb{C}$
We do have
$|z{|}^{2}=z\overline{z}$
and
$\overline{{z}_{1}+{z}_{2}}=\overline{{z}_{1}}+\overline{{z}_{2}}\phantom{\rule{0ex}{0ex}}\overline{{z}_{1}{z}_{2}}=\overline{{z}_{1}{z}_{2}}$
But if we take we find:
$|1-i×i|=|2|=2$
and
$|1+i×i|=|0|=0$

Vasquez

Expert

Step 1 $|\frac{a-ib}{a+bi}|$ $=\frac{|a-ib|}{|a+ib|}$ $=\frac{\sqrt{{a}^{2}+\left(-b{\right)}^{2}}}{\sqrt{{a}^{2}+{b}^{2}}}$ $=\underset{―}{1}$

alenahelenash

Expert

Step 1Yes, presuming a and b are real. Then this is the quotient of conjugates, which have the same magnitude, so the magnitude of the quotient is 1. The result is on the unit circle, at double the angle of the numerator.$|\frac{a-bi}{a+bi}|=\frac{|a-bi|}{|a+bi|}=\frac{\sqrt{{a}^{2}+{b}^{2}}}{\sqrt{{a}^{2}+{b}^{2}}}=1$If a and b are non-real complex numbers then in general this magnitude won't be 1.