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#1 2006-12-12 10:25:16

fangree01
Member
Registered: 2006-11-13
Posts: 11

Complex Numbers Help Plaese!!

Hi there,

I was wondering if someone could help with this question.

By putting z = x=iy (x, y ∈ R).  Solve the equation 2z² - z*² + 6z* + 9 = 0.  Were z* is the conjugate of z.

I can do a few of these but i am getting stuck with the 6x i am gettting out of the equation.

I would be grateful for any help.

Thanks

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#2 2006-12-12 11:38:24

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Complex Numbers Help Plaese!!

Let's start by replacing the z's with x's and y's then.

2z² - z*² + 6z* + 9 = 0.

2(x+iy)² - (x-iy)² + 6(x-iy) + 9 = 0 (z = x+iy, z* = x-iy).

Multiplying out: 2x² + 4xyi - 2y² - x² + 2xyi + y² + 6x - 6yi + 9 = 0 (i² = -1)

Simplifying: x² - y² + 6x + 9 + (6xy - 6y)i = 0

0 is the same as 0 + 0i, so we can split the equation into real and imaginary parts.

x² - y² + 6x + 9 = 0
6xy - 6y = 0

Sorry if you'd already done that bit. You commented on the 6x, which shows that you were on the right track. Ah well, it least it confirms that you were right up to that stage. smile

To continue, we must solve the 2 simultaneous equations we have here. Let's look at the imaginary one first because it looks easier.

6xy - 6y = 0
Rearrange: x = 6y/6y = y/y = 1 except when y = 0, and then x can take any value because if we substitute back then we'll always get 0 - 0 = 0, which works.

Now we can look at the real part.

The 6x was confusing you, but this equation is actually quite helpful because it involves the 6x in a perfect square: (x+3)² - y² = 0.
In general, you can always get rid of linear x and y terms by completing the square in the same way, but you'll sometimes get a remainder that you need to consider.

So, (x+3)² - y² = 0.
From before, we know that one of x=1 or y=0 must be true for a solution.

We can find the accompanying y to go with the x=1 by (1+3)² - y² = 0.
16 - y² = 0
y² = ± 4

Similarly, we can find the accompanying x to go with y=0.
(x+3)² - 0 = 0
(x+3)² = 0
x = -3.

So there are 3 solutions to the original equation:

z = -3
z = 1+4i
z = 1-4i


Why did the vector cross the road?
It wanted to be normal.

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