Find all the points where $f$ is analytic with $f(z)= \frac{z^2+1}{(3z-1)(z-i+1)}$.

Question

I start by expanding the denominator and separating the real and imaginary but get stuck when deciding what my $u$ and $v$ should be.

Thanks.

Answer 1

Avoid looking at real and imaginary parts. Sum and product of analytic functions are analytic. Quotient is analytic where the denominator is nonzero.

Answer 2

Hint. One may observe that, for $a \neq0$ and for $|z|<|a|$, we have

$$ z \mapsto \frac{1}{a+z}=\sum_{n=0}^\infty\frac{(-1)^nz^n}{a^{n+1}} \tag1 $$

and the considered function is analytic over $|z|<|a|$.

Now, by partial a fraction decomposition, one may obtain

$$ \frac{z^2+1}{(3z-1)(z-i+1)}= \frac{1}{3}-\frac{\frac{2}{5}-\frac{i}{5}}{1-i+z}+\frac{\frac{8}{45}+\frac{2i}{15}}{-1/3+z} \tag2 $$

then one may conclude using $(1)$.