# Complex and real forms of the Poisson integral formula

In my complex analysis book there is the expression $$\frac{1 - |z|^2}{|1 - \bar z e^{it}|^2}$$

and it says that when $z = re^{it}$, we can write the above expression as $$P_r(t) = \frac{1 - r^2}{1 - 2r\cos t + r^2} = \text{Re}\left( \frac{1 + z}{1 - z} \right)$$

I do not see where the $\cos t$ comes from though.

Isn't $\bar z = re^{-it}$, so the top is $1 - r^2$ and the bottom is $|1 - r|^2 = 1 - 2r + r^2$. I have not really figured out where the $\text{Re}(1 + z)/(1 - z)$ comes from either.

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By the way, expressions lacking an equals sign are not equations. – anon Jul 30 '12 at 8:29
Please provide book and page numbers if possible. There is something wrong with the expression you give for the Poisson kernel: there is a complex exponential there as usual, but its phase is (you say) the same as that of $z$. The phase should be a different variable independent of $z$. – anon Jul 30 '12 at 8:49
The book is complex variables by fisher. Page 285. I think I figured out what I misunderstood. I think the $t$ in the two expressions are actually a different variable. In complex analysis by gamelin (pg 276), the expression is different. – zrbecker Jul 30 '12 at 8:52

Here is a link a of the book. $P_r(t)$ is defined at equation (5) as $\frac{1-r^2}{1-2r\cos t+r^2}$. If $z=re^{it}$, then $$\frac{1+z}{1-z}=\frac{1+re^{it}}{1-re^{it}}=\frac{(1+re^{it})(1-re^{-it})}{(1-re^{it})(1-re^{-it})}=\frac{1-re^{-it}+re^{it}-r^2}{1-2r\cos t+r^2},$$ and the real part of $re^{-it}+re^{it}$ is $0$.
It helps us to write a formula for $f(re^{it})$, involving $P_r$.