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

- Thread starter Punch
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- Thread starter
- #1

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

- Feb 5, 2012

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Hi Punch,

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

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.

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I followed as you said but couldn't find how to extract the real part out...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.

- Feb 5, 2012

- 1,621

What did you get after multiplying by \(x-(y-2)i\) ?I followed as you said but couldn't find how to extract the real part out...

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- #5

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

But I do not see how i can use this

- Feb 5, 2012

- 1,621

\[Re(w)=\frac{x^2+x+y^2-2y}{x^2+y^2-4y+4}\]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

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 \(c\).