Operations with complex numbers to give real numbers

Question

If:

$|z|=|w|=1$
$1 + zw \neq 0$

Then $\dfrac{z+w}{1+zw}$ is real. How can prove that.

One thing, if you want to see whether $\frac{a}{b}$ is real (or purely imaginary) is to multiply with $\frac{\overline{b}}{\overline{b}}$. — Daniel Fischer, Jun 19 '14 at 17:56
See http://math.stackexchange.com/questions/427663/prove-if-z-w-1-and-1zw-neq-0-then-zw-over-1zw-is-a-real?rq=1 . — beep-boop, Jun 19 '14 at 17:57

score 5 · Accepted Answer · answered Jun 19 '14 at 17:56

5

Hint: Take the complex conjugate then multiply top and bottom by $zw$.

answered Jun 19 '14 at 17:56

André Nicolas

507,029

score 1 · Answer 2 · answered Jun 19 '14 at 17:57

Let $z=e^{ia}$ and $w=e^{ib}$ for some $a,b \in [0,2\pi)$. This takes care of condition 1. Then $1+zw \neq 0$ means $1+e^{i(a+b)} \neq 0$, which is the same as saying $a+b$ is not an odd multiple of $\pi$.

Now consider \begin{align*} \dfrac{z+w}{1+zw} & = \frac{e^{ia}+e^{ib}}{1+e^{i(a+b)}} \end{align*} Rationalize the denominator and see what happens.

Operations with complex numbers to give real numbers

2 Answers2