Animated Solution for Mathematics - Complex Numbers: Let $z$ be a complex number such that $|z+2|=1$ and $\text{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\text{Re}(\bar{z}+2)|$ is

Welcome, fellow traveler, to the fascinating world of complex numbers. Often, students view complex numbers as a dry collection of algebraic rules—$i^2=-1$, conjugate this, modulus that. But today, we are going to see them for what they truly are: a beautiful, geometric language.

We are looking at the equation $|z+2|=1$. In the complex plane, this is not just an equation; it is a story. It tells us that the distance between the point $z$ and the point $-2$ is always exactly $1$.

If you were to sketch this, you would immediately see a circle centered at $(-2,0)$ with a radius of $1$. This circle is our playground. Every point $z$ that satisfies this condition lives on this boundary.

Now, how do we navigate this circle? We could use Cartesian coordinates, setting $z=x+iy$, but that path is fraught with messy algebra. Instead, let's use the elegance of polar form.

Since the vector $z+2$ starts at the center $(-2,0)$ and ends at $z$ on the circle, its length is constant at $1$. We can represent this vector as $z+2=e^{i\theta }$, where $\theta$ is the angle the vector makes with the positive real axis.

This substitution is a game-changer. It transforms our complex variable $z$ into a simple angular parameter $\theta$. Suddenly, the entire problem becomes a dance of trigonometry.

Our goal is to analyze the expression $\text{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. This looks intimidating, but let's look closer.

We can rewrite the numerator $z+1$ as $(z+2)-1$. Because we already know exactly what $z+2$ is, we can split the fraction:

Now, substitute our polar form $z+2=e^{i\theta }$. The expression becomes $1-\frac{1}{e^{i\theta }}$, which is simply $1-e^{-i\theta }$. See how the complexity just melts away?

Now, we invoke the legendary Euler's formula: $e^{-i\theta }=\cos \theta -i\sin \theta$. Substituting this into our expression, we get $1-(\cos \theta -i\sin \theta )$.

Distributing the negative sign, we have $(1-\cos \theta )+i\sin \theta$. The problem tells us that the imaginary part of this expression is $\frac{1}{5}$.

Looking at our result, the imaginary part is clearly $\sin \theta$. Therefore, we have discovered that $\sin \theta =\frac{1}{5}$. This is the key that unlocks the final door.

We are asked to find $|\text{Re}(\bar{z}+2)|$. Let's analyze $\bar{z}+2$. We know $z+2=e^{i\theta }$. Taking the conjugate of both sides, we get $\overline{z+2}=\overline{e^{i\theta }}$, which simplifies to $\bar{z}+2=e^{-i\theta }$.

Expanding this using Euler's formula again, we get $\cos \theta -i\sin \theta$. The real part is simply $\cos \theta$. The problem asks for the absolute value of this real part, so we need $|\cos \theta |$.

We know $\sin \theta =\frac{1}{5}$. Using the fundamental identity ${\cos }^2\theta +{\sin }^2\theta =1$, we find:

Taking the square root, we get:

And there it is! The elegance of the result is a testament to the power of geometric thinking. You have navigated the circle, mastered the substitution, and arrived at the truth. The final answer is $\frac{2\sqrt{6}}{5}$.