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Complex Numbers - Finding Values


Date: 02/13/99 at 05:30:20
From: T. Wilson
Subject: Complex Numbers - Finding Values

Just a couple of questions I'm having difficulty with:

1) If (a + ib)^2 = 3 + 4i where a and b are real, find the values of 
   a and b.

2) Show that if a + bi is a root of a quadratic equation with real 
   coefficients, then the quadratic equation must have form              
   k(x^2 - 2ax + [a^2 + b^2] = 0, where K does not equal 0.

3) a + ai is a root of x^2 - 6x + b = 0.  Explain why b has two 
   possible values.  Find a in each case.  

Thanks.


Date: 02/13/99 at 08:53:06
From: Doctor Mitteldorf
Subject: Re: Complex Numbers - Finding Values

Dear Mr. Wilson,

The only trick here is not to get spooked by the word "imaginary." 
There's nothing difficult or mysterious here. You just follow the 
algebra in a straightforward way.

1) What is the square (a+ib)?  Multiply it out to get a^2 - b^2 + 2abi. 
That has to be equal to 3 + 4i. The only way to do this is to have the 
a^2+b^2 be the 3 and the 2ab be the 4. These are two equations with two 
unknowns. Solve them in the usual way.

2) Just keep the definitions straight, and translate what they're 
asking into algebra. Suppose your quadratic equation is kx^2+Bx+C = 0.  
Substitute x = a+bi into that, and you have k(a^2+b^2+2abi) + B(a+bi) + 
C = 0. They tell you that A and B are real, so the imaginary part of 
the equation must be zero: k*(2ab) + Bb = 0. Can you take it from here?

3) Substitute x = a+ai into the equation x^2 - 6x + b = 0 and make the 
one complex equation into two real equations by saying that both the 
imaginary and real parts are zero. Solve the two equations 
simultaneously for a and b, and see what happens.

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

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