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Given an entire function $f(z)$ we know that $$ \log f(z)=\log|f(z)|+i \arg f(z)$$

Let $f(z)=\frac{z-a}{z-\bar a}$, for complex number $a$. How to show that $\arg f(z)=\tan^{-1}(???)$ on the real line $\mathbb R$. (I don't know what is exactly inside $\tan^{-1}$)

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$\tan^{-1}$ is a notation for the inverse function of $\tan$, in other words $\arctan$. – AD. Jan 6 '12 at 21:27

How about first doing $\arg(x+iy)$ to see the arctan? Then, for your problem, write $f(z)$ in real and imaginary parts. Or maybe before the general case, write $z-a$ in real and imaginary parts.

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You can write $$f(z) = \frac{z-a}{z-\bar{a}} = \frac{(z-a)\overline{(z-\bar{a})}}{|z-\bar{a}|^2}.$$

Then, you express $\arg f(z)$ in a simplier form as $$\arg \frac{(z-a)\overline{(z-\bar{a})}}{|z-\bar{a}|^2} = \arg (z-a)\overline{(z-\bar{a})}.$$

Let $z = x+ iy$ and $a = p+iq$. Then, you can write $$(z-a)\overline{(z-\bar{a})} = ((p-x)^2-q^2+y^2) + i2q(p-x).$$ Therefore $$\arg f(z) = \arctan \dfrac{2q(p-x)}{(p-x)^2-q^2+y^2} $$

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