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# How do you find three cube roots of $1+i$ ?

Last updated date: 18th Sep 2024
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Hint:
To find the cube roots of $1+i$ in the standard $a+ib$ form where I is the imaginary part of the number and a and b are real numbers. Then we will write $1+i$ in polar form. The polar form is written as $z=r\left( \cos \theta +i\sin \theta \right)$ . Then we will apply De moivre’s theorem to find the cube roots.

We have been given $1+i$.
We have to find the three cube roots of $1+i$.
Now, first of all we need to write it in polar form
We know that polar form of any complex number $a+ib$ is written as $z=r\left( \cos \theta +i\sin \theta \right)$
Where $r=\sqrt{{{a}^{2}}+{{b}^{2}}}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{a}{b} \right)$ a and b are real numbers and i is the imaginary part.
Now, substituting the values we get
$\Rightarrow r=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}$ and
\begin{align} & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{1}{1} \right) \\ & \Rightarrow \theta ={{\tan }^{-1}}1 \\ & \Rightarrow \theta =\dfrac{\pi }{4} \\ \end{align}
Now, the polar form of $1+i$ will be
$\Rightarrow 1+i=\sqrt{2}\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)$
Now we have to find the cube root of $1+i={{\left( 1+i \right)}^{\dfrac{1}{3}}}$
We know that the ${{n}^{th}}$ root of the complex number is given by
${{r}^{\dfrac{1}{n}}}\left( \cos \dfrac{x+2\pi k}{n}+i\sin \dfrac{x+2\pi k}{n} \right)$ where k varies over the integer values from 0 to $n-1$
Therefore the cube root of the given expression is given as
$\Rightarrow {{\left( 1+i \right)}^{\dfrac{1}{3}}}={{\left( \sqrt{2} \right)}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{2n\pi +\dfrac{\pi }{4}}{3} \right)+i\sin \left( \dfrac{2n\pi +\dfrac{\pi }{4}}{3} \right) \right)$
Now, substituting the values $n=0,1,2$ we get three roots as
$\Rightarrow \left( \sqrt[6]{2} \right)\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right),\left( \sqrt[6]{2} \right)\left( \cos \left( \dfrac{3\pi }{4} \right)+i\sin \left( \dfrac{3\pi }{4} \right) \right)$ and $\left( \sqrt[6]{2} \right)\left( \cos \left( \dfrac{17\pi }{12} \right)+i\sin \left( \dfrac{17\pi }{12} \right) \right)$

Note:
while solving this question avoid basic calculation mistakes. To solve such types of questions students must have knowledge of complex numbers. Also, students must remember the De moivre’s theorem to solve this question. The De moivre’s theorem is valid for fractions also.