Question

Consider the complex numbers tex-formula"> and $w$ below:
Calculate $(z+w)^{9}$ in rectangular form.

Answer 1

Judging by the given polar plots, we have

$z = e^{i\,16\pi/12} = e^{i\,4\pi/3}$

$w = e^{i\,10\pi/12} = e^{i\,5\pi/6}$

Then

$z + w = e^{i\,4\pi/3} + e^{i\,5\pi/6} \\\\ ~~~~~~~~ = e^{i\,5\pi/6} \left(e^{i\,\pi/2} + 1\right) \\\\ ~~~~~~~~ = e^{i\,5\pi/6} (1 + i) \\\\ ~~~~~~~~ = (1 + i) w$

Compute the modulus and argument.

$|z + w| = |1+i| |w| = \sqrt2$

$\arg(z+w) = \arg(1+i) + \arg(w) = \dfrac\pi4 + \dfrac{5\pi}6 = \dfrac{13\pi}{12}$

or equivalently, -11π/12, to ensure the argument is between -π and π. So we have

$z + w = \sqrt2 \, e^{-i\,11\pi/12}$

Then by de Moivre's theorem,

$(z + w)^9 = \left(\sqrt2 \, e^{-i\,11\pi/12}\right)^9 \\\\ ~~~~~~~~ = \left(\sqrt2\right)^9\,e^{-i\,99\pi/12} \\\\ ~~~~~~~~ = 2^{9/2}\,e^{-i\,\pi/4} \\\\ ~~~~~~~~ = 2^{9/2} \left(\cos\left(\dfrac\pi4\right) - i \sin\left(\dfrac\pi4\right)\right) \\\\ ~~~~~~~~ = 2^{8/2} - i\,2^{8/2} = \boxed{16 (1 - i)}$