# Homework Help: Complex numbers. Imaginary part

1. Jan 14, 2004

### 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

-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: Jan 14, 2004
2. Jan 14, 2004

### himanshu121

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

i.e $$\frac{1}{x+iy}$$

What was/are ur thoughts???

3. Jan 14, 2004

### 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?

4. Jan 14, 2004

### himanshu121

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

5. Jan 14, 2004

### arcnets

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

6. Jan 14, 2004

### chroot

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

- Warren

7. Jan 15, 2004

### himanshu121

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

Last edited: Jan 15, 2004
8. Jan 15, 2004

### fannemel

$$\frac{Im_{z2}}{x^2 + y^2} = Im [\frac{1}{z1}]$$

Would that be any better?

For me that would equal:

$${\frac{-y}{x^2 + y^2} = Im [\frac{1}{z1}]$$
since $$Im_{z2} = -y$$

Last edited: Jan 15, 2004
9. Jan 17, 2004

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

10. Jan 17, 2004

### HallsofIvy

That is true because $$x^2+ y^2$$ 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 $$z_1$$ and $$z_2$$ by x+iy and x-iy right from the start:

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