Question

Consider the complex numbers
`z` 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) = \frac\pi4 + \frac{5\pi}6 = (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(\frac\pi4\right) - i \sin\left(\frac\pi4\right)\right) \\\\ ~~~~~~~~ = 2^(8/2) - i\,2^(8/2) = \boxed{16 (1 - i)}`