Question

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

Answer 1

Given:

The expression is:

`(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,

`(1)/(2)(\cos (72)+i\sin (72))^5`

Using De Moivre's theorem, we get

`=(1)/(2)(\cos (72* 5)+i\sin (72* 5))`

`=(1)/(2)(\cos (360)+i\sin (360))`

`=(1)/(2)(1+i(0))`

`=(1)/(2)+0i`

It is the
`a+bi` form of the given expression. Here,
`a=(1)/(2),\ b=0`.

Therefore, the required expression is
`(1)/(2)+0i`.