# Complex numbers equation with conjugate.

• April 29th 2011, 12:12 PM
TriForce
Complex numbers equation with conjugate.
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?
• April 29th 2011, 12:24 PM
topsquark
Quote:

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
• April 29th 2011, 12:29 PM
Plato
Quote:

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 $3a+5b=0~\&~(-3b+5a)=1$
• April 29th 2011, 12:33 PM
TriForce
Oh, thanks so much, i really need to practice more...
• April 29th 2011, 12:41 PM
Plato
Quote:

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.
• April 29th 2011, 01:52 PM
TriForce
Quote:

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

Could you please explain why you put real numbers = 0, and complex = 1, and where did complex i go?
• April 29th 2011, 02:12 PM
Plato
Quote:

Originally Posted by TriForce
Could you please explain why you put real numbers = 0, and complex = 1, and where did complex i go?

If $a+bi=3-4i$ then $a=3~\&~b=-4$.

So if $a+bi=i=0+i$ then $a=0~\&~b=1$.