I am doing homework and we just learned differentiability on complex functions so I am not sure how to check where is this function differentiable:
1/(z^5-1)
I tried with definition but I got stuck on lim((1/((z+h)^10-(z+h)^5))/h) as h goes to zero.
Could you help me a bit?
Here is a method using the limit definition of the derivative: $$\lim_{h\rightarrow0}\frac{\frac{1}{(z+h)^5-1}-\frac{1}{z^5-1}}{h}=\lim_{h\rightarrow0}\frac{z^5-1+1-(z+h)^5}{h\big((z+h)^5-1\big)(z^5-1)}$$ $$=\lim_{h\rightarrow0}-\frac{(z+h)^5-z^5}{h}\cdot\frac{1}{\big((z+h)^5-1\big)(z^5-1)}$$ $$=-\frac{d}{dz}(z^5)\lim_{h\rightarrow0}\frac{1}{\big((z+h)^5-1\big)(z^5-1)}$$ $$=-5z^4\lim_{h\rightarrow0}\frac{1}{(z^5-1)(h^5+5h^4z+10h^3z^2+10h^2z^3+5hz^4+z^5-1)}$$ $$=-\frac{5z^4}{(z^{5}-1)^2}$$ Note that I assumed the derivative of $z^5$ was known. If needed you can derive this for yourself again using the limit definition. This limit is defined for all $z\in\mathbb{C}$ except for when $z^5-1=0$. Can you solve for the five complex values for when the denominator is zero?