derlingasmh

2022-01-30

How do you find the power ${\left(-2+2i\right)}^{3}$ and express the result in rectangular form?

gekraamdbk

Expert

Step 1
We use polar complex first t o simplify down
Let $R{e}^{i\theta }=-2+2i$
$=2\sqrt{2}\left(-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$
$=2\sqrt{2}\left(-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$
We will find -ve value for $\mathrm{cos}\theta$ and +ve value for $\mathrm{sin}\theta$ in Q2 of the Argand diagram.
$⇒\theta =\frac{3\pi }{4}$
$⇒R{e}^{i\theta }=2\sqrt{2}{e}^{i\frac{3\pi }{4}}$
$⇒{\left(R{e}^{i\theta }\right)}^{3}={\left(2\sqrt{2}\right)}^{3}{e}^{i\frac{3×\pi }{4}}$
$=16\sqrt{2}{e}^{i\frac{\pi }{4}}$
$=16\left(1+i\right)$
clearly you could also use a binomial expansion

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