Question

How do I find the nth root of a complex number ?

Answer 1

Hint: nth root of a complex number means we will find the $\dfrac{1}{n}$ power of the complex number. We will use Euler’s formula to write the given complex number as a product of mod of complex number and e to the power argument of the complex number and i.

Complete step-by-step answer:
Let’s take the complex number as a + ib
By applying Euler’s formula we can write a + ib as $A{{e}^{ix}}$ where A is the mod of the complex number a + ib and x is the argument of the complex number
Argument of a + ib is the polar $\theta $ coordinate of the point ( a , b ) and the value of mod of the number is $\sqrt{{{a}^{2}}+{{b}^{2}}}$
After writing as $A{{e}^{ix}}$ we have to calculate ${{\left( A{{e}^{ix}} \right)}^{\dfrac{1}{n}}}$ we know that ${{\left( ab \right)}^{x}}={{a}^{x}}{{b}^{x}}$
So we can write ${{\left( A{{e}^{ix}} \right)}^{\dfrac{1}{n}}}={{A}^{\dfrac{1}{n}}}{{e}^{i\dfrac{x}{n}}}$
So the value of nth root of a + ib is equal to ${{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{2n}}}{{e}^{i\dfrac{x}{n}}}$
Again we can apply Euler’s formula
${{\left( a+ib \right)}^{\dfrac{1}{n}}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{2n}}}\left( \cos \dfrac{x}{n}+i\sin \dfrac{x}{n} \right)$
The RHS of the above equation is the nth root of complex number a + ib.

Note: The range of the argument is complex number is from $-\pi $ to $\pi $ where $-\pi $ is not included. Let’s take a complex number a + ib , if (a , b) is in the first quadrant the argument lies in the range $\left[ 0,\dfrac{\pi }{2} \right]$ , if it is second quadrant the argument lies in $\left[ \dfrac{\pi }{2},\pi \right]$, for third quadrant the range is $\left( -\pi ,-\dfrac{\pi }{2} \right]$ and for fourth quadrant the range is $\left[ -\dfrac{\pi }{2},0 \right]$ .