Complex numbers. Imaginary part

fannemel

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)?

himanshu121

They want the coefficient of i or Im(1/z)

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

What was/are ur thoughts???[8)]

fannemel

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?

himanshu121

No u can't change them that ways. It will defy all the properties of complex number

arcnets

fannemel,
I think your work is correct. Because, if 2 numbers are equal, then their imaginary parts are equal.

chroot

Yes, fannemel, you are entirely correct. Given z = x + iy, Im(1/z) is indeed -y / (x^2 + y^2).

- Warren

himanshu121

His Questions is
i.e.
[tex]\frac{Im_{z2}}{x^2 + y^2} = \frac{1}{Im_{z1}}[/tex]
Which is not true

fannemel

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]

ahmad

Yes that is absolutely correct. You get the imaginary part of the reciprocal of z1, not z1

HallsofIvy

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].