Complex numbers equation with conjugate.

**TriForce** · April 29th 2011, 12:12 PM

I'm having a bit problem solving the following equation:

c * conjugate(z) = i where c = 3 + 5i and z is a complex number (a + bi)

I tried following

conjugate(z) = i/c
But have no idea how to remove the conjugate from z.

Any ideas?

**topsquark** · April 29th 2011, 12:24 PM

Originally Posted by TriForce

I'm having a bit problem solving the following equation:

c * conjugate(z) = i where c = 3 + 5i and z is a complex number (a + bi)

I tried following

conjugate(z) = i/c
But have no idea how to remove the conjugate from z.

Any ideas?

The conjugate of the conjugate of z is z itself. So what is the conjugate of i/c?

-Dan

**Plato** · April 29th 2011, 12:29 PM

Originally Posted by TriForce

I'm having a bit problem solving the following equation:

c * conjugate(z) = i where c = 3 + 5i and z is a complex number (a + bi)

If $z=a+b\mathif{i}$ then $\overline{z}=a-b\mathif{i}$ .
So $(3+5\mathif{i})(a-b\mathif{i})=(3a+5b)+(-3b+5a)\mathif{i}$ .

So solve $3a+5b=0~\&~(-3b+5a)=1$

**TriForce** · April 29th 2011, 12:33 PM

Oh, thanks so much, i really need to practice more...

**Plato** · April 29th 2011, 12:41 PM

Originally Posted by TriForce

@topsquark Well that's kind of the problem, i can't find anything about conjugating a complex number fraction...

If you insist upon this: $\frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$ .

Then $\overline{~\frac{1}{z}~} = \frac{z}{{\left| z \right|^2 }}$ .

But I find the system approach clearer.