Key Concepts and Formulas

• Modulus of a Complex Number: For a complex number $z = x+iy$, its modulus is denoted by $|z|$ and is calculated as $|z| = \sqrt{x^2+y^2}$. Geometrically, $|z-c|=r$ represents all complex numbers $z$ that lie on a circle centered at $c$ with radius $r$ in the complex plane.
• Complex Conjugate: The conjugate of $z = x+iy$ is $\bar{z} = x-iy$. An important property is that $\operatorname{Re}(z) = \operatorname{Re}(\bar{z})$. Also, for any complex number $Z$, $|Z|^2 = Z\bar{Z}$.
• Division of Complex Numbers: To find the real or imaginary part of a quotient $\frac{z_1}{z_2}$, it is common practice to multiply the numerator and denominator by the conjugate of the denominator: $\frac{z_1}{z_2} = \frac{z_1 \overline{z_2}}{z_2 \overline{z_2}} = \frac{z_1 \overline{z_2}}{|z_2|^2}$.

Step-by-Step Solution

Step 1: Substitute to simplify the given conditions.

To simplify the repeated expression $z+2$, we introduce a new complex variable: $w = z+2$ We rewrite the given conditions in terms of $w$. This makes the algebra easier to handle.

1. The condition $|z+2|=1$ becomes: $|w|=1$ Let $w = a+ib$, where $a = \operatorname{Re}(w)$ and $b = \operatorname{Im}(w)$. Then: $a^2+b^2=1 \quad \text{(Equation 1)}$

2. For the condition $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$, we express $z+1$ in terms of $w$. Since $w = z+2$, we have $z = w-2$. Substituting $z=w-2$ into $z+1$: $z+1 = (w-2)+1 = w-1$ So, the second condition becomes: $\operatorname{Im}\left(\frac{w-1}{w}\right)=\frac{1}{5}$

Step 2: Evaluate the imaginary part of $\frac{w-1}{w}$.

Substitute $w = a+ib$ into the expression: $\frac{w-1}{w} = \frac{(a+ib)-1}{a+ib} = \frac{(a-1)+ib}{a+ib}$ Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, $\bar{w} = a-ib$: $\frac{((a-1)+ib)(a-ib)}{(a+ib)(a-ib)} = \frac{(a-1)a - i(a-1)b + iab + b^2}{a^2+b^2}$ Using Equation 1, $a^2+b^2=1$, the denominator simplifies to 1: $= \frac{a^2-a - iab + ib + iab + b^2}{1}$ Combine the real and imaginary terms: $= (a^2+b^2-a) + i(b)$ Substitute $a^2+b^2=1$: $= (1-a) + ib$ The imaginary part of $\frac{w-1}{w}$ is $b$.

Step 3: Determine the value of $b$ from the given imaginary part.

From the second condition, $\operatorname{Im}\left(\frac{w-1}{w}\right)=\frac{1}{5}$, and since we found that $\operatorname{Im}\left(\frac{w-1}{w}\right) = b$: $b = \frac{1}{5}$

Step 4: Find the value of $a$ using the equation $|w|=1$.

Using Equation 1, $a^2+b^2=1$, and the value of $b$: $a^2 + \left(\frac{1}{5}\right)^2 = 1$ $a^2 + \frac{1}{25} = 1$ $a^2 = 1 - \frac{1}{25} = \frac{24}{25}$ $a = \pm \sqrt{\frac{24}{25}} = \pm \frac{\sqrt{24}}{\sqrt{25}} = \pm \frac{2\sqrt{6}}{5}$

Step 5: Calculate $|\operatorname{Re}(\overline{z+2})|$.

We need to find $|\operatorname{Re}(\overline{z+2})|$. Recall $w = z+2$. So we want $|\operatorname{Re}(\bar{w})|$. Since $w=a+ib$, the conjugate of $w$ is $\bar{w} = a-ib$. The real part of $\bar{w}$ is $\operatorname{Re}(\bar{w}) = a$. Therefore, we need to find $|a|$. From Step 4, $a = \pm \frac{2\sqrt{6}}{5}$. $|a| = \left|\pm \frac{2\sqrt{6}}{5}\right| = \frac{2\sqrt{6}}{5}$

Common Mistakes & Tips

• Complex Conjugate: Remember that $\operatorname{Re}(z) = \operatorname{Re}(\bar{z})$. Confusing the real and imaginary parts after conjugation is a common mistake.
• Strategic Substitution: Substituting $w=z+2$ simplifies the algebra. Look for opportunities to do this.
• Rationalization: Be careful with the signs when rationalizing the denominator of a complex fraction.

Summary

By substituting $w=z+2$, we simplified the problem and used the properties of complex numbers, their modulus, and conjugates to find the value of $|\operatorname{Re}(\overline{z+2})|$. We found that $b = \frac{1}{5}$ and $a = \pm \frac{2\sqrt{6}}{5}$, therefore $|\operatorname{Re}(\overline{z+2})| = |a| = \frac{2\sqrt{6}}{5}$.

Final Answer

The final answer is $\boxed{\frac{2 \sqrt{6}}{5}}$, which corresponds to option (A).