Expand $(\sqrt{5}+i)^5$, where $i^2=-1$

Question

I'm getting:

$i^5+5 \sqrt{5} i^4+50 i^3+50 \sqrt{5} i^2+125 i+25 \sqrt{5}$ Is this right?

Answer 1

Hint $\,\ (\sqrt a + \sqrt b)^5 = f(a,b)\sqrt a + f(b,a)\sqrt b\,\ $ for $\,\ f(x,y) = x^2\! + 10xy+5 y^2\ $ exploits symmetry to shorten computation, and yields an independent check.

Answer 2

Another way:

If $a+ib = r e^{i\, \theta}$, $(a+ib)^n= r^n e^{n\,i\, \theta} =r^n (\cos(n\,i\, \theta)+i \sin(n\,i\, \theta))$.