# The value of the cubic root of $-i$

So this was the question given to us. $\left(\iota=\sqrt{-1}\right)$

Value(s) of $\left(-\iota\right)^{\dfrac{1}{3}}$ are

(A) $\dfrac{\sqrt{3}-\iota}{2}$

(B) $\dfrac{\sqrt{3}+\iota}{2}$

(C) $\dfrac{-\sqrt{3}-\iota}{4}$

(D) $\dfrac{-\sqrt{3}+\iota}{2}$

And the answer, we were told, is option (A). I agree with this option; it can be easily obtained by putting $\left(-\iota\right)$ in polar form. But, my question is, why can't we rewrite $\left(-\iota\right)$ as $\iota^{3}$, and hence say that $\left(-\iota\right)^{\dfrac{1}{3}} = \iota$ ?

• Oh okay. Makes sense. Thanks! Nov 21, 2015 at 5:25

You are right: $i$ is another solution of the equation $z^3=-i$. There is also a third one, $\dfrac{1+i\sqrt{3}}{2}$.
The general answer can be given as follows . $i^3=-i$ so we write $x^3=-i$ now we want three solutions where one is $i$ so for solutions we can write $cos{\frac{2kπ+\theta}{n}}+isin{\frac{2kπ+\theta}{n}}$ where $k=(0,1..,n-1)$ so here $n=3$ so $k=(0,1,2)$ now plug in the values and get the answer. Here $\theta=3π/2 ,n=3$ hope you can do the rest.
Since $$-i=0+(-i)=\cos(3\pi/2)+\sin(3\pi/2)i$$ then you'll have one of the below versions for $z$: $$z=\cos\left(\frac{2k\pi+3\pi/2}{3}\right)+i\sin\left(\frac{2k\pi+3\pi/2}{3}\right),~~k=0,1,2$$