# compute all values of z such that z^5 = -30 can you go step by step please, i know i have to use demoivre's formula, k = 0 through k = 4. please help

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You need to consider the number to the right side as a complex number, whose real part is -30 and imaginary part is 0.

You need to convert the algebraic form of the complex to the right into the polar form such that:

`z' = -30` (algebraic form)

`z' = |z'|(cos alpha + i sin alpha) ` (polar form)

`|z'| = sqrt((-30)^2 + 0^2) => |z'| = 30`

`tan alpha = 0/(-30) => tan alpha = 0 => alpha = 0^o`

`z' = 30(cos0^o + i sin 0^o)`

The problem provides the information that `z^5 = -30 => z = root(5)(-30)`

You need to substitute `z' = 30(cos 0^o + i sin 0^o)` for z' = -30 such that:

`z = root(5)(30(cos 0^o + i sin 0^o))`

You may write the fifth root as a fractional power such that:

`z = (30(cos 0^o + i sin 0^o))^(1/5)`

`z = root(5)(30)(cos 0^o + i sin 0^o)^(1/5)`

You need to use De Moivre's theorem such that:

`z = root(5)(30)(cos ((0+2kpi)/5) + i sin ((0+2kpi)/5))`

You need to substitute 0 for k such that:

`z_0 = root(5)(30)(cos ((0+0)/5) + i sin ((0+0)/5))`

`z_0 = root(5)(30)(cos 0 + i sin 0)`

`z_1 = root(5)(30)(cos ((0+2pi)/5) + i sin ((0+2pi)/5))`

`z_1 = root(5)(30)(cos ((2pi)/5) + i sin ((2pi)/5))`

`z_2 = root(5)(30)(cos ((0+4pi)/5) + i sin ((0+4pi)/5))`

`z_2 = root(5)(30)(cos ((4pi)/5) + i sin ((4pi)/5))`

`z_3 = root(5)(30)(cos ((0+6pi)/5) + i sin ((0+6pi)/5))`

`z_3 = root(5)(30)(cos ((6pi)/5) + i sin ((6pi)/5))`

`z_4 = root(5)(30)(cos ((0+8pi)/5) + i sin ((+8pi)/5))`

`z_4 = root(5)(30)(cos ((8pi)/5) + i sin ((8pi)/5))`

**Hence, evaluating the roots to the given complex equation yields `z_0 = root(5)(30)(cos 0 + i sin 0) ; z_1 = root(5)(30)(cos ((2pi)/5) + i sin ((2pi)/5)) ; z_2 = root(5)(30)(cos ((4pi)/5) + i sin ((4pi)/5)) ; z_3 = root(5)(30)(cos ((6pi)/5) + i sin ((6pi)/5)) ; z_4 = root(5)(30)(cos ((8pi)/5) + i sin ((8pi)/5)).` **

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