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NemiM [27]
3 years ago
7

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

Mathematics
1 answer:
PIT_PIT [208]3 years ago
8 0

Given:

The expression is:

\dfrac{1}{2}(\cos (72)+i\sin (72))^5

To find:

The a+bi form for the given expression.

Solution:

According to De Moivre's theorem,

(\cos \theta+i\sin \theta)^n=\cos (n\theta )+i\sin (n\theta )

We have,

\dfrac{1}{2}(\cos (72)+i\sin (72))^5

Using De Moivre's theorem, we get

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

=\dfrac{1}{2}(\cos (360)+i\sin (360))

=\dfrac{1}{2}(1+i(0))

=\dfrac{1}{2}+0i

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

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

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