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Let $z$ be a nonreal complex number such that $|z| = 1.$ Find the real part of $\frac{1}{1 - z}.$

 Nov 27, 2020

Best Answer 

 #1
avatar+9185 
+1

The question implies that the real part of  \(\frac{1}{1-z}\)  is the same for all nonreal complex values of  z  such that  \(|z| = 1\) . So assuming that's true, we can pick any nonreal complex  z  such that  \(|z| = 1\)  and find the real part of  \(\frac{1}{1-z}\) .

 

Let's pick  z  =  0 + 1i

 

Then...

 

\({\frac{1}{1-z}}\ =\ \frac{1}{1-(0+1i)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\cdot\frac{1+1i}{1+1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-i^2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-(-1)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{2}+\frac{1}{2}i \)

 

 

The real part is  \(\boxed{\frac12}\)

 

 

We can check a few more cases to see that it is 1/2 for those as well:

 

z = cis( pi/6 )

z = cis( pi/4 )

 

 

It seems like no matter what the angle is, the real part is  1/2  smiley

 Nov 27, 2020
 #1
avatar+9185 
+1
Best Answer

The question implies that the real part of  \(\frac{1}{1-z}\)  is the same for all nonreal complex values of  z  such that  \(|z| = 1\) . So assuming that's true, we can pick any nonreal complex  z  such that  \(|z| = 1\)  and find the real part of  \(\frac{1}{1-z}\) .

 

Let's pick  z  =  0 + 1i

 

Then...

 

\({\frac{1}{1-z}}\ =\ \frac{1}{1-(0+1i)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\cdot\frac{1+1i}{1+1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-i^2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-(-1)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{2}+\frac{1}{2}i \)

 

 

The real part is  \(\boxed{\frac12}\)

 

 

We can check a few more cases to see that it is 1/2 for those as well:

 

z = cis( pi/6 )

z = cis( pi/4 )

 

 

It seems like no matter what the angle is, the real part is  1/2  smiley

hectictar Nov 27, 2020
 #2
avatar+116126 
+1

Looks good, hectictar  !!!!

 

 

cool cool cool

CPhill  Nov 27, 2020
 #3
avatar
+2

Thank you so much for your help!

Guest Nov 28, 2020

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