Solving a Quick Complex Number Question: Finding z in 4z + z(bar) = 5 + 9i

[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)

 

[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

 

[Dick]

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

 

[Mentallic]

lurflurf, I can't see where you are headed with your approach.

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

[tex]4(z+\bar{z})=4z+4\bar{z} \neq 4z+\bar{z}[/tex]

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

 

[vorcil]

lurflurf, I can't see where you are headed with your approach.



[tex]4(z+\bar{z})=4z+4\bar{z} \neq 4z+\bar{z}[/tex]

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

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?

 

[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:

[tex]4z+\bar{z}=5+9i[/tex]

let z=a+ib

[tex]4a+4ib+a-ib=5+9i[/tex]

[tex]5a+3ib\equiv 5+9i[/tex]

Hence,
[tex]5a=5, a=1[/tex]
[tex]3ib=9i, b=3[/tex]

Therefore, [tex]z=1+3i[/tex]