# Thread: Complex numbers equation with conjugate.

1. ## 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?

2. 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

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

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

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

5. 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: $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$.

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

But I find the system approach clearer.

6. Originally Posted by Plato
So solve $\displaystyle 3a+5b=0~\&~(-3b+5a)=1$
Could you please explain why you put real numbers = 0, and complex = 1, and where did complex i go?

7. Originally Posted by TriForce
Could you please explain why you put real numbers = 0, and complex = 1, and where did complex i go?
If $\displaystyle a+bi=3-4i$ then $\displaystyle a=3~\&~b=-4$.

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