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### Square Root of i

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Date: 03/08/99 at 21:21:00
Subject: Square Root of i.

In my Honors Algebra II/Trigonometry class, we just completed a section
on complex numbers. One of my students asked me the following:

The square root of -1 is i, but what is the square root of i?

Can you help?
```

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Date: 03/09/99 at 08:31:37
From: Doctor Rick
Subject: Re: Square Root of i.

There are two complex numbers which, when squared, equal i. (The same
holds true for any number. But for real numbers, we arbitrarily say
that the positive root is THE square root. When the roots have
imaginary parts, any such choice would be even more arbitrary, and we
do not bother to choose one.)

The square roots of i are

(1 + i) * sqrt(2)/2 and
-(1 + i) * sqrt(2)/2

You can prove this just by squaring each number. To find them in the
first place, you can use Euler's formula

e^(i*t) = Cos(t)+i*Sin(t)

If t = pi/2, you get

e^(i*pi/2) = cos(pi/2)+i*sin(pi/2) = 0 + i*1 = i

The square root of this is

(e^(i*pi/2))^(1/2) = e^(i*pi/4)

and using Euler's formula again, we have

e^(i*pi/4) = 1/2 + i*1/2

If you set t = 5*pi/2, you again get e^(i*5pi/2) = i. Taking the square
root of this and using Euler's formula, you get the other root of i.

Euler's formula makes it easy to find powers and roots by working in
polar coordinates in the complex plane. Any number x + iy can be
written in terms of a radius r and angle theta (counterclockwise from
the x axis):

x + iy = re^(i*theta)
r = sqrt(x^2 + y^2)
theta = arctan(y/x)

Then, using Euler's formula,

(x+iy)^k = r^k e^(i*k*theta)
= r^k(cos(k*theta) + i*sin(k*theta))

Here are some related pages from our archives:

Square Roots in Complex Numbers
http://mathforum.org/dr.math/problems/jurd11.6.97.html

Proof of e^(iz) = cos(x) + isin(x)
http://mathforum.org/dr.math/problems/graf.4.7.97.html

Roots of Unity
http://mathforum.org/dr.math/problems/volante4.18.97.html

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Imaginary/Complex Numbers

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