Complex numbers. Imaginary part

  • Thread starter fannemel
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  • #1
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z1 = x + iy
z2 = x - iy
(Complex conjugate)

Find:
Im (1/z1)

This is what I have tried to do:

(1) z1*z2 = x^2 + y^2

(2) z2 / (x^2 + y^2) = 1 / z1


The answer is:
-y / (x^2 + y^2) = Im (1 / z1)

So my question is:
Can I change z2 to Im (z2) and z1 to Im (z1) in equation (2)?
 
Last edited:

Answers and Replies

  • #2
650
1
They want the coefficient of i or Im(1/z)

i.e [tex]\frac{1}{x+iy}[/tex]

What was/are ur thoughts???
 
  • #3
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oops, read another post and found that it would be wise to post my work. And i reckoned that no-one would have had the time to reply so i just edited my post.

But do my new post clarify anything?
 
  • #4
650
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No u can't change them that ways. It will defy all the properties of complex number
 
  • #5
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fannemel,
I think your work is correct. Because, if 2 numbers are equal, then their imaginary parts are equal.
 
  • #6
chroot
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Yes, fannemel, you are entirely correct. Given z = x + iy, Im(1/z) is indeed -y / (x^2 + y^2).

- Warren
 
  • #7
650
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Originally posted by chroot
Yes, fannemel, you are entirely correct. Given z = x + iy, Im(1/z) is indeed -y / (x^2 + y^2).

- Warren
His Questions is
So my question is:
Can I change z2 to Im (z2) and z1 to Im (z1) in equation (2)?
i.e.
[tex]\frac{Im_{z2}}{x^2 + y^2} = \frac{1}{Im_{z1}}[/tex]
Which is not true
 
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  • #8
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what about
[tex]\frac{Im_{z2}}{x^2 + y^2} = Im [\frac{1}{z1}][/tex]

Would that be any better?

For me that would equal:

[tex]{\frac{-y}{x^2 + y^2} = Im [\frac{1}{z1}][/tex]
since [tex]Im_{z2} = -y [/tex]
 
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  • #9
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Yes that is absolutely correct. You get the imaginary part of the reciprocal of z1, not z1
 
  • #10
HallsofIvy
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what about
[tex]\frac{Im{z_2}}{x^2 + y^2} = Im [\frac{1}{z1}][/tex]
Would that be any better?

For me that would equal:
[tex]{\frac{-y}{x^2 + y^2} = Im [\frac{1}{z1}][/tex]
since
[tex]Im{z_2} = -y [/tex]
That is true because [tex]x^2+ y^2[/tex] is a real number.
In general you cannot get the imaginary part of a number computed by a formula just by replacing each number in the formula by its imaginary part.

It would be far better for you to replace [tex]z_1[/tex] and [tex]z_2[/tex] by x+iy and x-iy right from the start:

[tex]\frac{1}{z_1}= \frac{1}{x+iy}[/tex]. Now multiply both numerator and denominator by x- iy to get [tex]\frac{(1)(x- iy)}{(x+iy)(x-iy)}= \frac{x- iy}{x^2+ y^2}= \(\frac{x}{x^2+y^2}\)-\(\frac{y}{x^2+y^2}\)i[/tex] so that it is obvious that the real part is [tex]\frac{x}{x^2+y^2}[/tex] and the imaginary part is [tex]\frac{-y}{x^2+y^2}[/tex].
 

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