My Coding > Mathematics > Complex numbers > Proper square root from -1

Proper square root from -1

We already know, that by definition, square root (\(\sqrt{-1}\)) is equal \(i\). But the question is, is this answer complete or not? In this tutorial, I will try to give you an answer to this interesting question.

We can solve this problem in two different ways. By squaring the equation, or by using the Euler form of a complex number. Let's try both of them.

Calculating of (\(\sqrt{-1}\)) by squaring

Calculating by squaring is not described in the video, so, I will try to be full in detail here.

We need to solve this equation with real \(a\) and \(b\):

\[\sqrt{-1} = a + bi\]

Let's square the left and right parts of this equation:

\[-1 = a^2 + 2abi - b^2\]

and now we need to make equal real and imaginary parts of this equation:

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

from the first line we know that either \(a\) or \(b\) should equal to \(0\). If \(b = 0\) then \(a^2 = -1\) and this cannot be solver for real values. Therefore we need to take \(a = 0\) and we will then have \(-b^2 = -1\) or \(b^2 = 1\) or \(b = \pm 1\).

Therefore:

\[\sqrt{-1} = \pm i\]

Calculating of (\(\sqrt{-1}\)) via Euler representation

This part is shown on video, so I will mention only main steps:

\[-1 = e^{i(\pi + 2\pi n)}\]

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

And now we can split our solution into two groups

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

Therefore:

\[\sqrt{-1} = \pm i\]

You can easily check, that this is correct y squaring both parts of this equality.

For more detailed explanation, please go to watch our video:


Published: 2023-05-11 02:56:35
Updated: 2023-05-11 03:01:44

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