Question

Use DeMoivre's Theorem to write the complex number in standard form.

Answer 1

Answer:

A

Step-by-step explanation:

First express $\sqrt{2}$ + i$\sqrt{2}$ in trig form

| z | = $\sqrt{(\sqrt{2})^2+(\sqrt{2})^2  }$ = $\sqrt{4}$ = 2

Θ = $tan^{-1}$( $\frac{\sqrt{2} }{\sqrt{2} }$) = $tan^{-1}$( 1) = $\frac{\pi }{4}$

Thus

$\sqrt{2}$ + i$\sqrt{2}$ = 2 [ cos$\frac{\pi }{4}$ + isin$\frac{\pi }{4}$ ]

Hence

($\sqrt{2}$ + i$\sqrt{2}$ )³

= 2³ [ cos ($\frac{\pi }{4}$ × 3) + isin($\frac{\pi }{4}$ × 3 ) ]

= 8 [ cos($\frac{3\pi }{4}$ ) + isin($\frac{3\pi }{4}$) ] → A

Answer 2

The standard form of the give complex number is

8 [ cos(\frac{3\pi }{4} ) + isin(\frac{3\pi }{4}) ]

We have given that,

First express  $\sqrt{2} + i\sqrt{2}$  in trig form

$| z | = \sqrt{(\sqrt{2})^2+(\sqrt{2})^2 } = \sqrt{4} = 2$

$\theta = tan^{-1}( \frac{\sqrt{2} }{\sqrt{2} }) = tan^{-1}( 1) = \frac{\pi }{4}$

Thus, $\sqrt{2} + i\sqrt{2} = 2 [ cos\frac{\pi }{4} + isin\frac{\pi }{4} ]$

What is the Demoivre's theore?

[ r(cos θ + i sin θ) ]^n = r^n(cos nθ + i sin nθ)

Hence, $(\sqrt{2} + i\sqrt{2} )^3$

$= 2^3 [ cos (\frac{\pi }{4} \times 3) + isin(\frac{\pi }{4} \times 3 ) ]$

$= 8 [ cos(\frac{3\pi }{4} ) + isin(\frac{3\pi }{4}) ] \implies A$

Thererefore the standard form of the give complex number is

8 [ cos(\frac{3\pi }{4} ) + isin(\frac{3\pi }{4}) ].

To learn more about the complex number visit:

https://brainly.com/question/10662770

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