I've gotten stuck on this problem: "Let f(z) be the branch to $z^{\dfrac {1}{3}}+z^{\dfrac {1}{4}}$, which is defined on the entire complex plane except on the negative imaginary axis, for which f(1)=0. Find f(-1)."
I started by writing $z^{\dfrac {1}{3}}+z^{\dfrac {1}{4}}$ as $e^{\dfrac {1}{3}\left( \ln \left| z\right| +i\theta \left( z\right) +i2n\pi \right) } + e^{\dfrac {1}{4}\left( \ln \left| z\right| +i\theta \left( z\right) +i2m\pi \right) }$ but when I try to solve f(1)=0 I just get the relationship between n and m, which means I won't get a specific branch. I am also not sure how to use the condition "not defined on the negative imaginary axis".
The answer is: f(-1)= $\dfrac {1-\sqrt {2}}{2}+i\left( \dfrac {\sqrt {3}-\sqrt {2}}{2}\right) $
Suggestions:
List the cube roots of $1$ and the fourth roots of $1$; for which choice(s) do you have $f(z) = z^{1/3} + z^{1/4}$ vanishing at $z = 1$?
If a branch of argument (polar angle) takes the value $\theta$ on the positive real axis, then the branch with a cut along the negative imaginary axis takes values in $(\theta - \frac{\pi}{2}, \theta + \frac{3\pi}{2})$, and you want to evaluate where the argument is $\theta + \pi$.