Problem and Analysis

The problem at hand requires us to prove that a specific expression involving complex numbers is a real number. The expression in question is given by:

$$frac{z_z}{z_z}$$
Where z and z' are complex numbers of modulus 1 and their sum is non-zero.

Solution Strategy

To solve this problem, we will first express the complex numbers z and z' in terms of their exponential forms. Then, we will simplify the given expression and analyze its properties.

Using Exponential Forms

We start by expressing z and z' using Euler's formula:

$$z e^{ia}$$

$$z' e^{ib}$$

where a and b are arbitrary real numbers.

Expression Simplification

Substituting these into the given expression, we get:

[frac{z_z}{z_z} frac{z_z cdot bar{z}bar{z'}}{z_z cdot bar{z}bar{z'}}]

(where (bar{z}) and (bar{z'}) are the complex conjugates of z and z', respectively).

Further simplifying:

[frac{z_z}{z_z} frac{z^2z'z^2bar{z}bar{z'}}{zbar{z'}^2} frac{z'bar{z}bar{z'}z^2}{bar{z}z^2} frac{bar{z}z'zbar{z'}}{zbar{z'}} frac{2operatorname{Re}(zbar{z'})}{zbar{z'}}]

This can be further simplified to:

[operatorname{Re}(zbar{z'})]

Since (operatorname{Re}(zbar{z'})) is always a real number and the modulus of z and z' is 1, the expression is indeed a real number.

Alternative Methods

Renaming and Numerical Experiments

An alternative method involves renaming the complex numbers to simplify the expression. Let:

$$z_1 z_2 1$$

Then, the given expression becomes:

[frac{z_1z_2}{z_1z_2}]

Hence, it is straightforward to see that the result is 1, which is a real number.

Direct Simplification

We can also directly simplify the expression by substituting:

$$z e^{ialpha}, z' e^{ibeta}$$

Then:

[frac{1e^{ialpha}e^{ibeta}}{e^{ialpha}e^{ibeta}} frac{1e^{i(alpha beta)}}{e^{ialpha}e^{ibeta}} cdot frac{e^{-ialpha}e^{-ibeta}}{e^{-ialpha}e^{-ibeta}} frac{e^{-ialpha}e^{-ibeta}e^{ibeta}e^{ialpha}}{1e^{i(alpha-beta)}e^{i(beta-alpha)}1} frac{2cosalpha2cosbeta}{22cos(alpha-beta)} frac{cosalphacosbeta}{1cos(alpha-beta)}]

In this form, the expression is clearly a real number, as it is the ratio of two real numbers.

Conclusion

The proof demonstrates that the expression involving complex numbers of modulus 1 is indeed a real number. By using the properties of complex conjugates and the exponential forms of complex numbers, we can efficiently show that the given expression always results in a real number.