Animated Solution for Mathematics - Complex Numbers: Let $z$ and $\omega$ be two complex numbers such that $|z|\le 1,|\omega |\le 1$ and $|z+i\omega |=|z-i\bar{\omega }|=2$ then $z$ equals

We are working within the complex plane where two complex numbers, $z$ and $\omega$, are constrained by the conditions $|z|\le 1$ and $|\omega |\le 1$. These numbers are restricted to the unit circle or its interior.

We are given two specific geometric constraints:

Consider the first condition $|z+i\omega |=2$. By the Triangle Inequality, we know that:

Since $|i|=1$, this simplifies to $|z|+|\omega |$. Given that the maximum value for both $|z|$ and $|\omega |$ is $1$, the maximum possible value for their sum is $1+1=2$.

Because the problem states that $|z+i\omega |$ is exactly $2$, the vectors $z$ and $i\omega$ must be perfectly aligned. This forces the magnitudes to their maximum values:

Since the vectors are collinear and point in the same direction, we can establish the relationship $z=k(i\omega )$ where $k=1$. This yields our first vital equation:

Now, consider the second condition: $|z-i\bar{\omega }|=2$. We can rewrite this as $|z+(-i\bar{\omega })|=2$.

Applying the same logic, since $|z|=1$ and $|-i\bar{\omega }|=|\bar{\omega }|=1$, the vectors must again be perfectly aligned. This leads to our second relation:

We now have a system of two equations: 1) $z=i\omega$ 2) $z=-i\bar{\omega }$

From the first equation, we isolate $\omega$ by multiplying both sides by $-i$:

Taking the conjugate of both sides, we find:

Substituting this expression for $\bar{\omega }$ into our second relation, we get:

Simplifying the right side using $i^2=-1$:

The condition $z=\bar{z}$ implies that $z$ has no imaginary part, meaning $z$ must be a purely real number.

We previously established that $|z|=1$. The only real numbers with a magnitude of $1$ are $1$ and $-1$.

Therefore, the possible values for $z$ are: