MHB Solve Complex Number IV Problem: Find Radius & Centre

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The discussion focuses on finding the radius and center of a circle represented by the complex number w in an Argand diagram, where the real part of w is zero. The expression for w is given as w=(z+1)/(z-2i), and participants are working through the manipulation of this equation to isolate the real part. After performing the necessary calculations, it is established that the condition Re(w)=0 leads to the equation x^2+x+y^2-2y=0. The goal is to transform this equation into the standard circle form (x-a)^2+(y-b)^2=r^2 to determine the circle's parameters. The discussion emphasizes the importance of correctly calculating and rearranging the equation to identify the center and radius of the circle.
Punch
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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
 
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Punch said:
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. :)
 
Sudharaka said:
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...
 
Punch said:
I followed as you said but couldn't find how to extract the real part out...

What did you get after multiplying by \(x-(y-2)i\) ?
 
Sudharaka said:
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
 
Punch said:
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 \(c\).
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

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