# Complex numbers. Imaginary part

• fannemel

#### 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) = I am (1 / z1)

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

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

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

What was/are ur thoughts?

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?

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

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

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

- Warren

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 I am (z2) and z1 to I am (z1) in equation (2)?

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

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

Would that be any better?

For me that would equal:

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

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

$$\frac{Im{z_2}}{x^2 + y^2} = I am [\frac{1}{z1}]$$
Would that be any better?

For me that would equal:
$${\frac{-y}{x^2 + y^2} = I am [\frac{1}{z1}]$$
since
$$Im{z_2} = -y$$

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 $$\frac{-y}{x^2+y^2}$$.