Square root of i

How to calculate the square root from \(i\)? This is pretty easy if you will use an exponential form of this complex number and do remember its periodicity. It is very important to remember that it will be more than one root.

By solving equiation

We want to find all complex numbers \(z\) that satisfy \(z^2 = i\). Writing \(z\) in the form \(a + bi\), where \(a\) and \(b\) are real numbers, we have \((a + bi)^2 = i\), which gives us:

\[\begin{aligned} (a + bi)^2 &= i \\ a^2 - b^2 + 2abi &= i \end{aligned}\]

Equating the real and imaginary parts, we get the system of equations:

\[\begin{aligned} a^2 - b^2 = 0 \\ 2ab = 1\end{aligned}\]

From the first equation, we have \(a^2 = b^2\), so \(a = \pm b\). Substituting into the second equation, we get \(2a^2 = 1\), so \(a = \pm \frac{1}{\sqrt{2}}\). Therefore, the solutions for \(z\) are:

\[\begin{aligned} z_1 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \\ z_2 &= -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \end{aligned}\]

Therefore, the two square roots of \(i\) are \(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\) and \(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\).

But this way is only suitable when we need to work with square roots. If we have higher order roots or fractional power, then this method became more complicated or even not applicable, therefore it is better to use exponential approach:

Via exponetial form

From Euler's formula we have \(i = e^{i(\frac{\pi}{2} + 2\pi*n)}\), so we can write:

\[\sqrt{i} = i^{\frac{1}{2}} = e^{\frac{1}{2}i(\frac{\pi}{2} + 2\pi*n)} = e^{i(\frac{\pi}{4} + \pi*n)}\]

Converting back to the classical form we will have two answers:

\[\sqrt{i} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}; -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\]

For graphical explanation please check this video:

Published: 2023-05-08 14:58:45
Updated: 2023-05-08 15:47:24