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

## Homework Statement

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

## 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?

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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!

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?
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!

vela
Staff Emeritus
Homework Helper

## Homework Statement

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

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

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!
ofcause!! oh how foolish i am :)