simplifying complex numbers

**StMartin** · 05-20-08, 08:58 AM

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

Simplify [itex](1+i\sqrt{2})^5-(1-i\sqrt{2})^5[/itex]

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

$(\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{\s qrt{6}}{3}))^5-(\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{-\sqrt{6}}{3}))^5<br />$
How will I get integer angle out of here?

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

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

**algebraic topology** · 05-21-08, 04:45 PM

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}$

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