# Integral Powers and Roots of a Complex Number

The square or the second power of a complex number $z$ is the product of $z$ by itself i.e. $z\cdot z$ and is denoted by $z^2$. Similarly, the cube or the third power of $z$, denoted by $z^3$, is defined by $z^3=z^2\cdot z$.

In general, for any positive integer $n$, the $n^{\text{th}}$ power of a complex number $z$ is defined by $z^n=z^{n-1}\cdot z\;\;\;\text{and,}\;\;\; z^0=1$

Obviously, the $n^{\text{th}}$ power of a complex number is also a complex number. Let us denote it by $w$, so that $z^n=w$

Conversely, if $w≠0$, and if $n$ is a positive integer, then any complex number $z$ whose $n^{\text{th}}$ power is $w$, is known as the $n^{\text{th}}$ root of $w$.

In other words, any complex number $z$ such that $z^n=w,\;\; w≠0,\;\; n=1,2,3,…$ is known as the $n^{\text{th}}$ root of $w$. The $n^{\text{th}}$ root of $w$ is usually denoted by $w^{\frac{1}{n}}\;\;\;\text{or}\;\;\; \sqrt[n]{w}$

The computation of the $n^{\text{th}}$ power and the $n^{\text{th}}$ root of a complex number may be carried out with comparative case with the help of a theorem known as De Moivre’s Theorem, which uses the polar (or trigonometric) form of the complex number.

## nth Root of a Complex Number

Let $z=r(\cos\theta+i\sin\theta)$ be a complex number. Let $w=R(\cos\phi+i\sin\phi)$ be the $n^{\text{th}}$ root of the given complex number $z$. Then,

$\begin{array}{l} & w^n &= z \\ \Rightarrow & [R(\cos\phi+i\sin\phi)]^n &= r(\cos\theta+i\sin\theta) \\ \Rightarrow & R^n(\cos n\phi+i\sin n\phi) &= r(\cos\theta+i\sin\theta) \end{array}$

Since the two complex numbers are equal, so $R^n=r\Rightarrow R=r^{\frac{1}{n}}=\sqrt[n]{r}$ and, $\cos n\phi+i\sin n\phi=\cos\theta+i\sin\theta$

$\begin{array}{l} \therefore & \cos n\phi = cos\theta &\text{and}& \sin n\phi=\sin\theta \\ \Rightarrow & n\phi=2k.π+\theta &\text{or,}& k.360°+\theta \\ \therefore & \phi=\frac{k.360°+\theta}{n} \end{array}$

$\begin{array}{l} \therefore w &= R(\cos\phi+i\sin\phi) \\ &= \sqrt[n]{r}\left[\cos\frac{k.360°+\theta}{n} +i\sin\frac{k.360°+\theta}{n}\right] \end{array}$

For $k=0,1,2,3,…,n-1$, we get $n$ different roots of $z$. For other integral value of $k$, it will give the root obtained earlier i.e. roots will be repeated.

Thus, the $k^{\text{th}}$ root of $z$ denoted by $z_k$ is given by $z_k=\sqrt[k]{r}\left[\cos\frac{k.360°+\theta}{n} +i\sin\frac{k.360°+\theta}{n}\right]$ $\text{where,}\;\; k=0,1,2,3,…,n-1$

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