Square Roots Of Complex Numbers

Date: 02/22/99 at 00:47:05
From: Max Newman
Subject: Square Roots Of Complex Numbers

Find the square roots of 5-12i.

Date: 02/22/99 at 05:02:43
From: Doctor Mitteldorf
Subject: Re: Square Roots Of Complex Numbers

There are two ways to think about this problem. The straightforward but 
tedious way is this: suppose some number (a+bi) satisfies the condition 
that multiplied by itself it gives 5-12i. Well, you can actually 
multiply out a+bi times itself, and get a^2-b^2 for the real part and 
2ab for the imaginary part. Therefore, a^2-b^2 = 5 and 2ab = -12. If you 
solve two equations with two unknowns, you get answers for a and b. 
Note that you are working with a quadratic equation, so you will have 
two solutions. There is always a positive and a negative square root of 
every number, even a complex one.

Here is a clever way to think about the question that may be a bit over 
your head for now. A complex number x+iy can be represented as a point 
in the x-y plane. However, it is often useful to represent that same 
point with polar coordinates r and t. "r" is the distance from the 
origin (sqrt(x^2+y^2)), which is the modulus of the complex number. 
"t" is the angle from the x-axis, satisfying tan(t)=y/x. When you take 
the square root of the number, you are taking the square root of the 
modulus, so r -> sqrt(r). Interestingly (I will not prove this today) 
the angle t is cut in half: t -> t/2. 

Another way to take the square root of a complex number is to use the 
tangent half-angle formula, or actually find the angle, halve it, and 
take the tangent. Of course, you will probably want to express the 
answer as real + imaginary parts when you are done, so you will reverse 
the process. In your example, it works like this: 

   r = sqrt(5^2+(-12)^2) = 13
 
so the modulus of the square root is sqrt(13). tan(t) = -12/5, so 
t = -1.176 radians. The t for the square root is -0.588 radians. This 
means that your answer for the square root is 

   sqrt(13)*(cos(-.588) + i*sin(-.588)) 
   = 3 - 2i

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