Let $Z_1$ and $Z_2$ be complex numbers such that $\frac{3Z_1}{4Z_2}$ is purely imaginary.What is $\left| \frac{Z_1 + Z_2}{Z_1 - Z_2} \right|$ equal to ?

Question

Prepp Team · Accepted Answer

Complex Number Magnitude Calculation

The problem asks for the value of $\left| \frac{Z_1 + Z_2}{Z_1 - Z_2} \right|$ given that $\frac{3Z_1}{4Z_2}$ is purely imaginary.

Condition Analysis

A complex number $w$ is purely imaginary if its real part is zero, which implies $w = - \bar{w}$ (for $w \neq 0$).

Given that $\frac{3Z_1}{4Z_2}$ is purely imaginary:

$ \frac{3Z_1}{4Z_2} = - \overline{\left(\frac{3Z_1}{4Z_2}\right)} $ $ \frac{3Z_1}{4Z_2} = - \frac{3\bar{Z_1}}{4\bar{Z_2}} $

Simplifying this equation by canceling out the common terms $\frac{3}{4}$:

$ \frac{Z_1}{Z_2} = - \frac{\bar{Z_1}}{\bar{Z_2}} $

This can be rewritten as:

$ \frac{Z_1}{Z_2} = - \overline{\left(\frac{Z_1}{Z_2}\right)} $

This shows that the ratio $\frac{Z_1}{Z_2}$ is itself a purely imaginary number. Let's represent this ratio as $ik$, where $k$ is a non-zero real number.

$ \frac{Z_1}{Z_2} = ik, \quad k \in \mathbb{R}, k \neq 0 $

Magnitude Calculation

Now, we need to find the magnitude of the expression $\frac{Z_1 + Z_2}{Z_1 - Z_2}$. Divide both the numerator and the denominator by $Z_2$ (assuming $Z_2 \neq 0$):

$ \frac{Z_1 + Z_2}{Z_1 - Z_2} = \frac{\frac{Z_1}{Z_2} + \frac{Z_2}{Z_2}}{\frac{Z_1}{Z_2} - \frac{Z_2}{Z_2}} = \frac{\frac{Z_1}{Z_2} + 1}{\frac{Z_1}{Z_2} - 1} $

Substitute $\frac{Z_1}{Z_2} = ik$ into the expression:

$ \frac{ik + 1}{ik - 1} = \frac{1 + ik}{-1 + ik} $

Finally, calculate the magnitude of this complex number:

$ \left| \frac{1 + ik}{-1 + ik} \right| = \frac{|1 + ik|}{|-1 + ik|} $

The magnitude of a complex number $a+bi$ is $\sqrt{a^2 + b^2}$. Therefore:

$ \frac{|1 + ik|}{|-1 + ik|} = \frac{\sqrt{1^2 + k^2}}{\sqrt{(-1)^2 + k^2}} = \frac{\sqrt{1 + k^2}}{\sqrt{1 + k^2}} $ $ \frac{\sqrt{1 + k^2}}{\sqrt{1 + k^2}} = 1 $

Thus, the magnitude $\left| \frac{Z_1 + Z_2}{Z_1 - Z_2} \right|$ is equal to 1.