Question

How do you find the fourth roots of $i$ ?

Answer 1

Hint:We will first start by mentioning De Moivre’s Theorem. Then apply the theorem, and note all the values of $n$ for which we will solve. Then evaluate all the values for different values of $n$ and hence, evaluate the fourth roots of $i$.

Complete step by step answer:
Here we will start by using the De Moivre’s Theorem.
According to the theorem,
If $z = r{e^{i\theta }} = r(\cos \theta + i\sin \theta )$

Then ${z^n} = {r^n}{e^{i \times n\theta }} = r(\cos n\theta + i\sin n\theta )$

As $i = \cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)$ can also be written as $i = \cos \left( {2n\pi + \dfrac{\pi }{2}} \right) + i\sin \left( {2n\pi + \dfrac{\pi }{2}} \right)$

$\sqrt[4]{i} = {i^{\dfrac{1}{4}}} = \cos \left( {\dfrac{{2n\pi }}{4} + \dfrac{\pi }{8}} \right) + i\sin
\left( {\dfrac{{2n\pi }}{4} + \dfrac{\pi }{8}} \right)$ or $ = \cos \left( {\dfrac{{n\pi }}{2} + \dfrac{\pi
}{8}} \right) + i\sin \left( {\dfrac{{n\pi }}{2} + \dfrac{\pi }{8}} \right)$

Note that $n = 0,1,2,3\,$ and after $n = 3$ it will repeat.

This will give us fourth roots of $i$, which are

\[
= \cos (\dfrac{\pi }{8}) + i\sin (\dfrac{\pi }{8}) \\
= \cos (\dfrac{\pi }{2} + \dfrac{\pi }{8}) + i\sin (\dfrac{\pi }{2} + \dfrac{\pi }{8}) \\
= - i\sin (\dfrac{\pi }{8}) + \cos (\dfrac{\pi }{8}) \\
= \cos (\pi + \dfrac{\pi }{8}) + i\sin (\pi + \dfrac{\pi }{8}) \\
= - \cos (\dfrac{\pi }{8}) - i\sin (\dfrac{\pi }{8}) \\
\]

And

\[
\cos (\dfrac{{3\pi }}{2} + \dfrac{\pi }{8}) + i\sin (\dfrac{{3\pi }}{2} + \dfrac{\pi }{8}) \\
= i\sin (\dfrac{\pi }{8}) - \cos (\dfrac{\pi }{8}) \\
\]

And to get exact values we ca use
\[\sin (\dfrac{\pi }{8}) = \dfrac{{\sqrt {2 - \sqrt 2 } }}{2} = 0.3827\] and
\[\cos (\dfrac{\pi }{8}) = \dfrac{{\sqrt {2 + \sqrt 2 } }}{2} = 0.9239\]

Therefore, the fourth roots of $i$ are $0.9239 + 0.3827i,\, - 0.3827 + 0.9239i,\, - 0.9239 - 0.3827i$ and $ + 0.3827 - 0.9239i$.

Additional Information: Complex number is a number that can be expressed in the form of $a + ib$, where $a$ and $b$ are real numbers, and $i$ represents the imaginary unit, satisfying the equation ${i^2} = - 1$.

 Because no real number satisfies this equation, $i$ is called an imaginary number. Complex numbers allow solutions to certain equations that have no solutions in real numbers.

The idea is to extend the real numbers with an intermediate $i$ which is also called an imaginary unit taken to satisfy the relation ${i^2} = - 1$, so that solutions to equations like the preceding one can be found.

Geometrically, complex numbers extend the concept of the one-dimensional number line to the two-dimensional complex plane, by using the horizontal axis for the real part and vertical axis for the imaginary part.

Note: While applying the De Moivre’s Theorem make sure you are taking proper values. Also, remember that the value of ${i^2}$ is $ - 1$. When evaluating different values make sure you evaluate them along with their respective signs.