How can I put the equation \frac{a+ib}{1+a-ib} into the form a+bi?

[Fellowroot]

Homework Statement

Use the Definition Re(z1)=Re(z2), Im(z1)=Im(z2)to solve each equation for z=a+bi.

$\frac{z}{1+\bar{z}}$=3+4i

Homework Equations

Sec 1.1 #42 from Complex Analysis 2nd ed from Dennis Zill

The Attempt at a Solution

I have solved several similar problems like this one in my text but I'm getting stuck on this one part.

The goal is to say:

$\frac{z}{1+\bar{z}}$=$\frac{a+ib}{1+a-ib}$

and put the right hand side of this equation into a real part and an imaginary part and equate the real and imaginary parts to the original one given.

So in short how to I put $\frac{a+ib}{1+a-ib}$ into a+bi form?

I have tried many conjugates but none have worked

Thanks

 

[Char. Limit]

It might help if you carried out the division directly, i.e. remember that:

$$\frac{a+ib}{c+id} = \frac{ac+bd}{c^2+d^2} + i \frac{bc - ad}{c^2 + d^2}$$

Just let a+1=c and b=d for your division.

 

[ehild]

So in short how to I put $\frac{a+ib}{1+a-ib}$ into a+bi form?

That is not needed. Just substitute z=a+ib for z in the original equation, multiply both sides with the denominator, and compare both the real and imaginary parts on each sides.

ehild

 

[Fellowroot]

That is not needed. Just substitute z=a+ib for z in the original equation, multiply both sides with the denominator, and compare both the real and imaginary parts on each sides.

ehild

Thank you ehild, I got it with your advice. For some reason I just forgot that the real and imaginary parts can have both a's and b's in it.

 

[HallsofIvy]

Another way to solve
$$\frac{a+ ib}{1+a- ib}= 3+ 4i$$
is to multiply both sides by 1+ a- ib:
a+ ib= (3+ 4i)(1+ a- ib).

Multiply the right side out and equate real and imaginary parts.