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Complex Numbers in Second Degree Equation

Date: 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
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Associated Topics:
High School Imaginary/Complex Numbers

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