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### The Fourth Root of -1

```
Date: 03/27/98 at 09:30:33
From: Thomas W. Sweeny
Subject: Complex Numbers

Can you provide me with some information on how to find the square
root of -1. I believe it is the square root of two over two plus the
square root of two over two times i. But... how to get that?

Thanks. The question came up in my technical algebra 2 high school
class.
```

```
Date: 03/27/98 at 12:23:09
From: Doctor Rob
Subject: Re: Complex Numbers

I assume you actually mean sqrt(i), not sqrt(-1) = i.

Assume it has the form a + b*i, with a and b real numbers. Then

(a+b*i)^2 = i
a^2 + 2*a*b*i + b^2*i^2 = i
a^2 - b^2 = (1 - 2*a*b)*i

Now the left side is real, and so the right side must also be real.
This can only happen if 1 - 2*a*b = 0, and so a^2 - b^2 = 0.  Thus
we have to solve two simultaneous equations for a and b:

2*a*b = 1
a^2 - b^2 = 0

The last equation can be rewritten:

(a - b)*(a + b) = 0

So:

a = b     or    a = -b

which means:

2*b^2 = 1  or  -2*b^2 = 1

Since b is real, the last equation cannot be correct, so a = b, and

2*b^2 = 1
b^2 = 1/2
a = b = sqrt(2)/2  or  a = b = -sqrt(2)/2

sqrt(i) = sqrt(2)/2 + i*sqrt(2)/2
sqrt(i) = -sqrt(2)/2 - i*sqrt(2)/2

-Doctor Rob, The Math Forum
```

```
Date: 03/27/98 at 17:20:31
From: Thomas W. Sweeny
Subject: Re: Complex Numbers

Bummer . . . I was careless and probably in a hurry! :-(
I meant to ask about the fourth root of -1. Sorry. Perhaps you could
respond to that. And thank you for the quick response. My students
is a small class, and I think I have piqued their interest and they
are developing mathematical confidence. They really showed interest
in this problem and were wondering how our CAS program DERIVE and our
Scientific Notebook program were able to arrive at the answer.

Thanks again.

Tom Sweeny
Berlin High School
Berlin, NH
```

```
Date: 03/27/98 at 18:18:02
From: Doctor Anthony
Subject: Re: Complex Numbers

Once you go beyond the square root it is better to use DeMoivre's
theorem for finding say 4th roots. This theorem states that:

[cos(x) + i.sin(x)]^n =  cos(nx) + i.sin(nx)

-1 =  cos((2k+1)pi) + i.sin((2k+1)pi)     k = any integer

Then taking the 4th root of each side:

(-1)^(1/4) = cos((2k+1)pi/4) + i.sin((2k+1)pi/4)

There will be 4 4th roots found by taking k = 0, 1, 2, 3

k=0    z1 =  cos(pi/4) + i.sin(pi/4)
k=1    z2 =  cos(3pi/4) + i.sin(3pi/4)
k=2    z3 =  cos(5pi/4) + i.sin(5pi/4)
k=3    z4 =  cos(7pi/4) + i.sin(7pi/4)

If you plotted these values on the Argand diagram you would have the
four points on the unit circle at 45 degrees, 135 degrees, 225 degrees
and 315 degrees to the positive x axis.

-Doctor Anthony,  The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 03/27/98 at 19:08:41
From: Thomas W. Sweeny
Subject: Re: Complex Numbers

Thanks a lot. You are great!
```
Associated Topics:
High School Imaginary/Complex Numbers

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