Question

If z=-1-sq root 3i and z^6=a+bi then a = ___ and b= ___

Answer 1

a=64 and b=0

Explanation:

`z=-1-√(3)i`

use
`z=e^(ix)=cosx+isinx`

Now we find out x using given z

`z=-1-√(3)i`

Multiply and divide the right side by -2

`z=-2((1)/(2) +(√(3))/(2))`

We know cos(pi/3) = 1/2

sin(pi/3)= sqrt(3)/2

`z=-2((1)/(2) +(√(3))/(2))`

`z=-2(cos((\pi)/(3))+sin((\pi)/(3))`

compare the above equation with use
`z=e^(ix)=cosx+isinx`

x= pi/3

so
`z=-2e^{(\pi)/(3)i}`

`z^6=64(e^{(\pi)/(3)i})^6`

`z^6=64e^{(6\pi)/(3)i}`

`z^6=64e^(2\pi*i)`

Plug in 2pi in the z equation

`z=64e^(2\pi*i)=64(cos(2\pi)+isin(2\pi))`

So z= 64(1 + i(0))

z=64

So a= 64 and b=0