Question

How do you find the ${n^{th}}$ roots of a complex number in polar form ?

Answer 1

To find the root of a complex number in polar form, we will use Euler’s method. First convert the complex in Euler form and then take ${n^{th}}$ root of it, because that's what we have been asked for. And then after taking ${n^{th}}$ root, convert it into polar form. The Euler form of a complex number is written as,$a + ib = A{e^{i\theta }} = A(\cos \theta + i\sin \theta )$.Where the first expression is the general form of a complex number second is the Euler form and third is the polar form. Also, where $A\;{\text{and}}\;\theta$ are mod and argument of complex numbers respectively.

Complete step by step answer:
In order to find the ${n^{th}}$ roots of a complex number in polar form, let us consider a complex number $z = a + ib$ .We can express this in Euler form as follows
$z = A{e^{i\theta }},\;{\text{where}}\;A\;{\text{and}}\;\theta$ are mod and argument of complex number respectively.

Now, as we have asked to find the ${n^{th}}$ roots of a complex number in polar form, we will first take its ${n^{th}}$ root and then convert it into polar form.Taking ${n^{th}}$ root both sides of the above equation,
${z^{\dfrac{1}{n}}} = {\left( {A{e^{i\theta }}} \right)^{\dfrac{1}{n}}}$
Using distributive property of exponent over multiplication, we will get
${z^{\dfrac{1}{n}}} = {A^{\dfrac{1}{n}}}{e^{\dfrac{{i\theta }}{n}}}$
So we get the ${n^{th}}$ root, now we will convert it into polar form,
We know that $A{e^{i\theta }} = A(\cos \theta + i\sin \theta )$, using this we will get
${z^{\dfrac{1}{n}}} = {A^{\dfrac{1}{n}}}(\cos \dfrac{\theta }{n} + i\sin \dfrac{\theta }{n})$
Writing $r$ in place of $A$
$\therefore {z^{\dfrac{1}{n}}} = {r^{\dfrac{1}{n}}}(\cos \dfrac{\theta }{n} + i\sin \dfrac{\theta }{n})$

So, ${r^{\dfrac{1}{n}}}(\cos \dfrac{\theta }{n} + i\sin \dfrac{\theta }{n})$ is the polar form of the ${n^{th}}$ roots of a complex number, where $r = \sqrt {{a^2} + {b^2}} \;{\text{and}}\;\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$

Note: We can solve or find directly the ${n^{th}}$ roots of a complex number in polar form with the help of De Moivre’s Theorem that gives the direct formula for computing the powers of a complex numbers. For any complex number $z = a + ib$ and integer $n$ it is given as,
${z^n} = {r^n}(\cos n\theta + i\sin n\theta ),\;{\text{where}}\;r = \sqrt {{a^2} + {b^2}} \;{\text{and}}\;\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$