# Express the equation `z^ 4 = −2 + 2sqrt3i` in polar form. (solve the equation, expressing your answers in polar form.) `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))` .

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