Square Roots Of Complex NumbersDate: 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/ |
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