# Express the roots of unity in standard for a+bi. Determine z and the four fourth roots of z if one fourth root of z is -2 - 2i.

*print*Print*list*Cite

According to DeMoivre Theorem, the 4th root of a number

`z=r(costheta+isintheta)`

is given by `z_k=root(4)(r)(cos[(theta+2kPi)/4]+isin[(theta+2kPi)/4])`

Looking at the -2-2i=2(-1-i) this means that the sin and cos are equal, hence theta is either Pi/4 or Pi/4+Pi (1st and 3rd quadrant sin and cos have the same sign)

Since we both sin and cos are negative in the given problem then theta has to equal 5Pi.

`root(4)(r)[cos((5Pi)/4)+isin((5Pi)/4)]=-2-2i =>`

`root(4)(r)[-sqrt(2)/2-isqrt(2)/2]=-2-2i =>`

`root(4)(r)*sqrt(2)/2=2 =>`

`root(4)(r)=4/sqrt(2)=4sqrt(2)/2=2sqrt(2)=>`

`r=[2sqrt(2)]^4=2^4*sqrt(2)^4=16*4=64`

Thus `z=64(cos5Pi+isin5Pi)=64(-1+i*0)=-64`

The other roots can be obtained by adding 2Pi, 4Pi, and 6Pi to 5Pi. I will work one for you.

`z_2=2sqrt2[cos((7Pi)/4)+isin((7Pi)/4)]=`

`2sqrt(2)[-sqrt(2)/2+isqrt(2)/2)=-2+2i`

`z_3=2+2i`

`z_4=2-2i`

(To help you see geometrically, 7Pi/4 falls in the 4th quadrant, 9Pi/4=2Pi+Pi/4 falls in the 1st, and 11Pi/4=2Pi+3Pi/4 fall in the 2nd quadrant)