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Square Root of i


Date: 03/30/2001 at 14:22:15
From: Javier Guido
Subject: Value of i

Hello Dr. Math:

Our college's algebra teacher asked us to find the square root of i. 
I applied the properties of exponents and got (-1)^(1/4). I checked 
the result on my TI83 and got a different result.

My result: i ^(1/2) = ((-1)^1/2)^1/2 = (-1)^1/4
TI83: ((2)^1/2)/2 + i((2)^1/2)/2

I am intrigued by the outcome. I think the exponents' properties do 
not work all the time for imaginary numbers and I wonder why. Can you 
explain why, and how can I find the square root of i?


Date: 03/30/2001 at 15:48:17
From: Doctor Rob
Subject: Re: Value of i

Thanks for writing to Ask Dr. Math, Javier.

When you were asked to find sqrt(i), your teacher meant you to express
it in the form a + b*i, where a and b are real numbers.  (-1)^(1/4)
is not in that form. The answer from your TI-83 is correct. You can
verify it by squaring and simplifying to see that the result is i.

The laws of exponents that you learned work in some situations and
not others. Actually, you have seen this before: if the base is zero, 
you cannot have a negative exponent, because of the prohibition
against dividing by zero. They work if the base is zero and the
exponent is positive. The case of 0^0 is problematic, but usually it 
is best to make this equal to 1. That describes what happens when the 
base is zero. 

Now consider the case where the base is nonzero.

The laws of exponents work if the exponent is an integer, whatever the 
base. They work if the base is a positive real number and the exponent 
is real. They do NOT work if the base is a negative real number or 
complex and the exponent is irrational or fractional. (At least, they 
do not work in quite the same way.) This last is the situation above:  
the base i is complex and the exponent 1/2 is fractional. The reasons 
that they don't work, and ways of understanding why, and how to deal 
with this situation, are the subject of some study that you might well 
see in a second course in algebra, or even in trigonometry.  It boils 
down to this.

Every nonzero complex number can be expressed in the form

   r*[cos(t) + i*sin(t)],

for some real numbers r > 0 and t.  r is called the absolute value of
the complex number, and t is called its argument. In the example at
hand, i can be so expressed with r = 1 and t = Pi/2, 5*Pi/2, 9*Pi/2,
and so on. Then it is a theorem that

   [cos(t) + i*sin(t)]^n = cos(n*t) + i*sin(n*t).

This is called DeMoivre's Theorem. Proving it for nonnegative integers 
n is not too hard (using trigonometry and mathematical induction), 
and that implies its truth for negative integers and all rational 
numbers n. Thus if you want to find i^(1/2), you use r = 1, t = Pi/2, 
and n = 1/2, to see that

   i^(1/2) = cos(Pi/4) + i*sin(Pi/4) = sqrt(2)/2 + i*sqrt(2)/2

If you use the argument t = 5*Pi/2 instead, you get

   i^(1/2) = cos(5*Pi/4) + i*sin(5*Pi/4) = -sqrt(2)/2 - i*sqrt(2)/2

It should not be a surprise to you that i has two square roots, and
that each is the negative of the other.  All the other arguments of i
will give you one or the other of these two answers.

The same ideas can be used to find roots and powers of other complex
numbers, realizing that there are n distinct nth roots.

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

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