# How does one get the form of a circle out of this equation?

1. Aug 11, 2012

### Keshroom

1. The problem statement, all variables and given/known data

Sketch modulus((z+1)/(2z+3))=1 on the complex plane where z=x+iy

2. Relevant equations

3. The attempt at a solution
I know it is a circle but i need help simplifying the equation into the form of a circle.

i'm stuck at
0= 3x^2 + 3y^2 + 10x + 8

I usually complete the square on these type of problems but it has a 3x^2. What can i do to get it to the form of a circle?

2. Aug 11, 2012

### Millennial

If you use Latex, this looks neater:
$$\left| \frac{z+1}{2z+3} \right| = 1$$
Play around with the absolute value:
$$\frac{|z+1|}{|2z+3|} = 1$$
This means that $|z+1|=|2z+3|$.
Now substituting a+bi for z, we get $|(a+1)+bi|=|(2a+3)+2bi|$.
Using the definition of the absolute value of a complex number, we then obtain
$$a^2+2a+b^2+1=4a^2+12a+9+4b^2$$
Rearranging yields
$$3a^2+10a+3b^2+8=0$$
This is a quadratic equation which you can solve to get the imaginary part of z in terms of its real part. Then, plug in values for the real part and get the imaginary part values!

3. Aug 11, 2012

### DeIdeal

If you want to complete the square but can't because of the 3x2 there, what can you do to remove the 3? Divide by it!

4. Aug 11, 2012

### vela

Staff Emeritus
That's what you want to do. What specifically is it about completing the square that is a problem for you?

5. Aug 11, 2012

### Keshroom

ofcause!! oh how foolish i am :)