For questions 16 - 19, write each expression in the standard form for the complex number a + bi. 1/2(cos(72)+isin(72))^5

Respuesta :

Given:

The expression is:

[tex]\dfrac{1}{2}(\cos (72)+i\sin (72))^5[/tex]

To find:

The [tex]a+bi[/tex] form for the given expression.

Solution:

According to De Moivre's theorem,

[tex](\cos \theta+i\sin \theta)^n=\cos (n\theta )+i\sin (n\theta )[/tex]

We have,

[tex]\dfrac{1}{2}(\cos (72)+i\sin (72))^5[/tex]

Using De Moivre's theorem, we get

[tex]=\dfrac{1}{2}(\cos (72\times 5)+i\sin (72\times 5))[/tex]

[tex]=\dfrac{1}{2}(\cos (360)+i\sin (360))[/tex]

[tex]=\dfrac{1}{2}(1+i(0))[/tex]

[tex]=\dfrac{1}{2}+0i[/tex]

It is the [tex]a+bi[/tex] form of the given expression. Here, [tex]a=\dfrac{1}{2},\ b=0[/tex].

Therefore, the required expression is [tex]\dfrac{1}{2}+0i[/tex].

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