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complex number IV

Punch

New member
Jan 29, 2012
23
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
Feb 5, 2012
1,621
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
Jan 29, 2012
23
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
Feb 5, 2012
1,621
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
Jan 29, 2012
23
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
Feb 5, 2012
1,621
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 \(c\).