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# Intermediate Algebra Basic Complex Question

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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

#1
+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

Nov 27, 2020

#1
+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

hectictar Nov 27, 2020
#2
+116126
+1

Looks good, hectictar  !!!!

CPhill  Nov 27, 2020
#3
+2

Thank you so much for your help!

Guest Nov 28, 2020