# Quick complex number question

1. Oct 14, 2009

### vorcil

4z + z(bar) = 5 + 9i
then z = [solve for this]

I don't know how to re arrange this equation

(5+9i)/4 = z+z(bar)

2. Oct 14, 2009

### lurflurf

z+z(bar)=2Re(z)
but written
4z + z(bar) = 5 + 9i
is not the same as
(5+9i)/4 = z+z(bar)
but
4(z + z(bar) )= 5 + 9i
is

3. Oct 14, 2009

### Dick

Just write z=a+bi and separate it into real and imaginary parts.

4. Oct 14, 2009

### Mentallic

$$4(z+\bar{z})=4z+4\bar{z} \neq 4z+\bar{z}$$

vorcil take Dick's approach. Just remember that $\bar{z}$ is the complex conjugate of z, thus if z=a+ib then $\bar{z}$=a-ib

5. Oct 15, 2009

### vorcil

4z + z(bar) = 5 + 9i

4z(re) + z(bar)(re) = 5
4z + zbar = 5, z(re) = 1 since zbar(re) = z(re)

4z(im) + z(bar)(im) = 9i
4x - x = 9
z(im) = 3i

z = 1 + 3i
checking
4(1+3i) + (1-3i) = 5+9i

is that right?

6. Oct 15, 2009

### Mentallic

Yes that is correct, except just remember to be sure to re-read the question so you answer exactly what it asked. "solve for z"
So then just at the end write: z=1+3i

Also, just another similar method which you might find simpler and this usually makes a larger variety of questions more simple and easy to understand:

$$4z+\bar{z}=5+9i$$

let z=a+ib

$$4a+4ib+a-ib=5+9i$$

$$5a+3ib\equiv 5+9i$$

Hence,
$$5a=5, a=1$$
$$3ib=9i, b=3$$

Therefore, $$z=1+3i$$