How to calculate $\partial_x(x+i y)^{a+ib}$

Question

$a,b$ are constant. How to calculate $\partial_x(x+i y)^{a+ib}$ ?

What I try : $$ x+iy=\sqrt{x^2+y^2}e^{i(\arctan \frac{y}{x})}=e^{\ln\sqrt{x^2+y^2} +i(\arctan \frac{y}{x})} $$ So $$ (x+i y)^{a+ib}=e^{(a+ib)(\ln\sqrt{x^2+y^2} +i(\arctan \frac{y}{x}))}=e^{a\ln\sqrt{x^2+y^2}-b\arctan \frac{y}{x}+i(a\arctan \frac{y}{x}+b\ln\sqrt{x^2+y^2})} $$ So $$ \partial_x(x+i y)^{a+ib} =\partial_x e^{a\ln\sqrt{x^2+y^2}-b\arctan \frac{y}{x}+i(a\arctan \frac{y}{x}+b\ln\sqrt{x^2+y^2})} \\ =...... $$ At the last line , I think it is too complex in fact , so , whether there is a good way to calculate it ?

Answer 1

We can treat $y$ as constant since we are only interested in the $x$-derivative.

What you have is $f(x)^a f(x)^{ib}$ for $f(x)=x+iy$.

Use the product rule to separate the $f(x)^a$ and $ f(x)^{ib}$ terms.

$\partial _x \ f(x)^a = a f(x)^{a-1} f'(x)$ using the chain rule.

$\partial _x \ f(x)^{bi} = \partial _x \ (f(x)^b)^i$ is a bit harder. We should be certain what complex exponents mean before we start. . . .

For any number $X$ we have $X^i = \exp (i \log(x)) $

That gives $\partial _x \ (f(x)^b)^i = \partial _x \ \exp(i(f(x)^b)$.

That can be calculated using how the complex exponential is equal to its derivative, and the chain rule.

Finally put it al back together. You will be left with something simple enough.