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### Exploring i

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Date: 03/16/98 at 04:21:12
From: Matthew Lilley
Subject: i^0 and i^2

I have been reading about imaginary numbers on the Internet as I am
interested in Fractals, and was wondering whether i^0 is 1. I have
asked my maths teachers at school, and none of them know, but they
all seem to think it isn't one.

Also, on your page on imaginary numbers, you say that i raised to any
power but 0 is 1. Aren't i^2 and i^6 both equal to +1?

i = SQR(-1)
i*i should = -1.

In fractals (i.e. z = z^2+c for the Mandelbrot set) it does.

Thanks,
Matthew
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Date: 03/16/98 at 06:12:24
From: Doctor Mitteldorf
Subject: Re: i^0 and i^2

Dear Matthew,

Come to us for information any time - but it would be well if you
could find someone locally to mentor you in math.

Yes i^0 is 1. There are many ways to see why it is that anything
raised to the 0 power must be one. Here's one: remember the way that
exponents add when you multiply numbers and subtract when you divide
them. This principle tells us that a^(1-1) = a^1/a^1. In other words,
express the number zero as (1-1). Any number a raised to the (1-1)
power is a divided by a, which I'm sure you can believe is always 1.

Imaginary numbers are no exception to the rules of algebra. In fact,
since they're just "inventions," all we know about imaginary numbers
must be derived from the rules of algebra that can be discovered and
tested for real numbers.

i^2 = -1 by definition, so:

i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i

...and so on, in cycles of four. In fact, when you plot complex
numbers on a plane, with the real part on the x axis and the imaginary
part on the y axis, multiplication by i corresponds exactly to a
rotation through 90 degrees to the left. Four such rotations get you
back to where you started, corresponding to the fact that i^4 = 1.

-Doctor Mitteldorf,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Imaginary/Complex Numbers

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