Complex Analysis and Finding Derivatives

Question

Find the 6th derivative of $f(z) = exp((−1 + i\sqrt{3})z)$.

Not sure how to proceed. I know I can expand

$(−1 + i\sqrt{3})z= -x-yi+ix\sqrt3 -y \sqrt3$.

After this step, I know I could break up the exponential by the Laws of exponents.

Answer 1

Note sure why you would do that. The derivative of $f(z)=e^{az}$, is $f'(z)=ae^{az}$. So $$ f^{(6)}(z)=(-1+i\sqrt 3)^6\,e^{(-1+i\sqrt 3)z}. $$ If you want to make this just a tad simpler, $$ (-1+i\sqrt 3)^6=2^6(\cos \frac{2\pi}3+i\sin\frac{2\pi}3)^6 =64(\cos 4\pi+i\sin4\pi)=64. $$ So $$ f^{(6)}(z)=64\,e^{(-1+i\sqrt 3)z}. $$