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### Complex Numbers - Finding Values

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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.
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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
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/
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Associated Topics:
High School Imaginary/Complex Numbers

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