Complex Numbers in Second Degree EquationDate: 9/15/96 at 15:4:53 From: Anonymous Subject: Complex Numbers in Second Degree Equation Solve the equation z^2+(4-2i)z-8i = 0 How I tried to solve it: z = -(4-2i)/2 +- sqr( ( (4-2i)/2)^2 + 8i) ... and after some simplification: z = -2+i +- sqr(3+4i) But I don't know how to solve the square root of 3+4i. Is it possible? Have I simplified wrong? Or am I supposed to solve the problem in a different way? Yours sincerely, Anders Ekdahl Date: 9/16/96 at 1:1:57 From: Doctor Pete Subject: Re: Complex Numbers in Second Degree Equation So far your simplification is correct. You want to find the square root of 3+4I. Here's how you can figure it out: Say some complex number x+yI = Sqrt[3+4I]. So we want to solve for x, y. Square both sides. Then 3+4I = (x+yI)^2 = x^2 + 2xyI - y^2 = (x^2 - y^2) + 2xyI . Equating real and imaginary parts, we find x^2 - y^2 = 3 2xy = 4 So xy = 2. Can you guess a solution in integers? Well, take x=2 and y=1. Then xy = 2, and x^2 - y^2 = 4 - 1 = 3. So Sqrt[3+4I] = 2+I. But notice that there's another solution, namely Sqrt[3+4I] = -(2+I). Substituting this back into our equation, x = -2+I [+-] Sqrt[3+4I] = -2+I [+-] (2+I) <-- notice the +- took care of the = {2I, -4} two possible square roots. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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