+0

# I need help on this problem (pls)

0
101
1

The complex numbers z and w satisfy that the magnitude of z = the magnitude of w = 1 and zw is not equal to 1

Prove that  the following fraction is a real number.

$$\frac{z + w}{zw + 1}$$

(I also proved before that the magnitude of z is equal to 1/z and the same is true for w. I have a hint that is; a complex number is real if and only if it is equal to its own conjugate.)

Thank you so much, this is a great community and I would love to help in the future!

Apr 5, 2023

### 1+0 Answers

#1
0

We can write the given fraction as follows:

(z + w)/(zw + 1) = (z + w)/((z - w)(1 + zw))

Since the magnitude of z and w are both 1, we know that ∣z−w∣=1. This means that z−w is a complex number with magnitude 1.

We also know that zw is not equal to 1. This means that zw is a complex number with magnitude 1 that is not equal to 1.

Therefore, 1+zw is a complex number with magnitude 1.

Since the denominator of the given fraction is a complex number with magnitude 1, we know that the fraction is a real number.

Alternatively, we can prove that the fraction is a real number by using the following steps:

Let z=a+bi and w=c+di, where a, b, c, and d are real numbers.

Then, zw=(a+bi)(c+di)=(ac−bd)+(ad+bc)i.

Since zw is not equal to 1, we know that ac−bd neq 0.

Therefore, we can divide the numerator and denominator of the given fraction by i, which gives us the following fraction:

(z + w)/(zw + 1) = ((a + bi) + (c + di))/((ac - bd) + (ad + bc)i) = (a + c)/(ac - bd) + (b + d)/(ad + bc)i.

Since ac−bd neq 0, we know that the fraction above is a real number.

Apr 5, 2023