Answer:
See proof below
Step-by-step explanation:
Let [tex]r=\frac{w+z}{1+wz}[/tex]. If w=-z, then r=0 and r is real. Suppose that w≠-z, that is, r≠0.
Remember this useful identity: if x is a complex number then [tex]x\bar{x}=|x|^2[/tex] where [tex]\bar{x}[/tex] is the conjugate of x.
Now, using the properties of the conjugate (the conjugate of the sum(product) of two numbers is the sum(product) of the conjugates):
[tex]\frac{r}{\bar{r}}=\frac{w+z}{1+wz} \left(\frac{1+\bar{w}\bar{z}}{\bar{w}+\bar{z}}{\right)[/tex]
=[tex]\frac{(w+z)(1+\bar{w}\bar{z})}{(1+wz)(\bar{w}+\bar{z})}=\frac{w+z+w\bar{w}\bar{z}+z\bar{z}\bar{w}}{\bar{w}+\bar{z}+\bar{w}wz+\bar{z}zw}=\frac{w+z+w+|w|^2\bar{z}+|z|^2\bar{w}}{\bar{w}+\bar{z}+|w|^2z+|z|^2w}=\frac{w+z+\bar{z}+\bar{w}}{\bar{w}+\bar{z}+z+w}=1[/tex]
Thus [tex]\frac{r}{\bar{r}}=1[/tex]. From this, [tex]r=\bar{r}[/tex]. A complex number is real if and only if it is equal to its conjugate, therefore r is real.
