# complex number IV

#### Punch

##### New member
A complex number is represented by the point P in an Argand diagram. If the real part of the complex number w=\frac{z+1}{z-2i} (z not 2i) is zero, show that the locus of P is a circle and find the radius and centre of the circle.

I have a problem manipulating w to find the real part of w

#### Sudharaka

##### Well-known member
MHB Math Helper
A complex number is represented by the point P in an Argand diagram. If the real part of the complex number w=\frac{z+1}{z-2i} (z not 2i) is zero, show that the locus of P is a circle and find the radius and centre of the circle.

I have a problem manipulating w to find the real part of w
Hi Punch,

Let $$z=x+yi$$ where $$x,y\in\Re$$.

$w=\frac{z+1}{z-2i}=\frac{(x+1)+yi}{x+(y-2)i}$

Now multiply both the numerator and the denominator by $$x-(y-2)i$$. Hope you can continue.

#### Punch

##### New member
Hi Punch,

Let $$z=x+yi$$ where $$x,y\in\Re$$.

$w=\frac{z+1}{z-2i}=\frac{(x+1)+yi}{x+(y-2)i}$

Now multiply both the numerator and the denominator by $$x-(y-2)i$$. Hope you can continue.
I followed as you said but couldn't find how to extract the real part out...

#### Sudharaka

##### Well-known member
MHB Math Helper
I followed as you said but couldn't find how to extract the real part out...
What did you get after multiplying by $$x-(y-2)i$$ ?

#### Punch

##### New member
What did you get after multiplying by $$x-(y-2)i$$ ?
With regards to the real part, \frac{x^2+x+y^2+2y}{x^2+y^2-2y+4}=0

But I do not see how i can use this

#### Sudharaka

##### Well-known member
MHB Math Helper
With regards to the real part, $\frac{x^2+x+y^2+2y}{x^2+y^2-2y+4}=0$

But I do not see how i can use this
$Re(w)=\frac{x^2+x+y^2-2y}{x^2+y^2-4y+4}$

As you may see, you have calculated the real part incorrectly. Check your calculation again.

Since, $$Re(w)=0\Rightarrow x^2+x+y^2-2y=0$$

Now can you try to transform this equation into the form $$(x-a)^2+(y-b)^2=r^2$$. The complex number $$z$$ is on this circle. All you got to do is find $$a, b$$ and $$c$$.