1. The problem statement, all variables and given/known data

Simplify (1+i\sqrt{2})^5-(1-i\sqrt{2})^5

2. Relevant equations

z=a+bi

z=r(cos\varphi+isin\varphi)

tg\varphi=\frac{b}{a}

r=\sqrt{a^2+b^2}

3. The attempt at a solution

(\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{\s qrt{6}}{3}))^5-(\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{-\sqrt{6}}{3}))^5

How will I get integer angle out of here?

arccos\frac{\sqrt{3}}{3} \approx 54.74^\circ

arcsin\frac{\sqrt{-6}}{3} \approx -54.74^\circ

The best thing to do is just to expand the whole thing.

\(1+\mathrm{i}\sqrt{2}\)^5=1+\mathrm{i}\(5\sqrt{2} \)-20-\mathrm{i}\(20\sqrt{2}\)+20+\mathrm{i}(4\sqrt{2})= 1-11sqrt{2}\,\mathrm{i}

\(1-\mathrm{i}\sqrt{2}\)^5=1+\mathrm{i}\(-5\sqrt{2}\)-20-\mathrm{i}\(-20\sqrt{2}\)+20+\mathrm{i}\(-4\sqrt{2}\)=1+11sqrt{2}\,\mathrm{i}

\therefore\ \(1+\mathrm{i}\sqrt{2}\)^5-\(1-\mathrm{i}\sqrt{2}\)^5=-22\sqrt{2}\,\mathrm{i}