`z^4 = -2+2sqrt3 i`

The polar form of complex number `x + yi` is `r ( cos theta + i sin theta)`

where `r = sqrt(x^2+y^2) ` and `theta = tan^(-1) y/x` .

Base on the given complex number, `x=-2` and `y= - 2sqrt3` . So,

`r =((-2)^2+(2sqrt3)^2)...

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`z^4 = -2+2sqrt3 i`

The polar form of complex number `x + yi` is `r ( cos theta + i sin theta)`

where `r = sqrt(x^2+y^2) ` and `theta = tan^(-1) y/x` .

Base on the given complex number, `x=-2` and `y= - 2sqrt3` . So,

`r =((-2)^2+(2sqrt3)^2) = sqrt(4+12)=sqrt16=4`

And `theta` lies at the second quadrant since x is negative and the y is positive.

`theta = tan^(-1) (-2/(2sqrt3) )= tan^(-1) (-1/(sqrt3))=tan^(-1) (-sqrt3/3) = (5pi)/6`

Substituting the valuef of r and `theta` to the polar form yields:

`2 ( cos ((5pi)/6) + i sin ((5pi)/6 ))`

**Hence, the polar form of `z^4=-2 +2sqrt3 i ` is `2 (cos ((5pi)/6) + isin((5pi)/6))` .**