Question

If $a>0$ and $z = \frac{(1+i)^2}{a - i}$ ($i = \sqrt{-1}$) has magnitude $\frac{2}{\sqrt{5}}$, then $\overline{z}$ is

• A. $-\frac{2}{5}-\frac{4}{5}i$
• B. $-\frac{2}{5}+\frac{4}{5}i$
• C. $\frac{2}{5}-\frac{4}{5}i$
• D. $\frac{2}{5}+\frac{4}{5}i$

Hint:

Algebra Tip:Memorize that $(1+i)^2 = 2i$ and $(1-i)^2 = -2i$. This saves valuable time in complex number calculations.Properties of modulus can also be used: $|z| = \frac{|(1+i)^2|}{|a-i|} = \frac{(\sqrt{2})^2}{\sqrt{a^2+1}} = \frac{2}{\sqrt{a^2+1}}$. This skips the rationalization step entirely!

Answer 1

The Correct Option is A

Concept: Complex Numbers - Algebraic Operations, Modulus, and Conjugate.

Step 1: Simplify the numerator of the complex number $z$. Expand $(1+i)^{2} = 1^{2} + 2(1)(i) + i^{2}$. Since $i^{2} = -1$, this becomes $1 + 2i - 1 = 2i$. Thus, $z = \frac{2i}{a-i}$.

Step 2: Rationalize the denominator to express $z$ in standard form. Multiply the numerator and denominator by the conjugate of the denominator, $(a+i)$: $z = \frac{2i(a+i)}{(a-i)(a+i)} = \frac{2ai + 2i^{2}}{a^{2} - i^{2}} = \frac{2ai - 2}{a^{2} + 1} = \frac{-2}{a^{2}+1} + i\frac{2a}{a^{2}+1}$.

Step 3: Calculate the modulus $|z|$ and solve for $a$. The modulus is given by $\sqrt{\text{Re}(z)^2 + \text{Im}(z)^2}$. $|z| = \sqrt{\left(\frac{-2}{a^2+1}\right)^2 + \left(\frac{2a}{a^2+1}\right)^2} = \sqrt{\frac{4 + 4a^2}{(a^2+1)^2}} = \sqrt{\frac{4(1+a^2)}{(a^2+1)^2}} = \sqrt{\frac{4}{a^2+1}} = \frac{2}{\sqrt{a^2+1}}$.

Step 4: Equate to the given magnitude and find $a$. We are given that $|z| = \frac{2}{\sqrt{5}}$. So, $\frac{2}{\sqrt{a^2+1}} = \frac{2}{\sqrt{5}}$. This implies $a^2+1 = 5$, or $a^2 = 4$. Since the problem states $a>0$, we must choose $a = 2$.

Step 5: Substitute $a$ back into $z$ and find the conjugate $\overline{z}$. Substitute $a=2$ into our standard form expression from
Step 2: $z = \frac{-2}{(2)^2+1} + i\frac{2(2)}{(2)^2+1} = \frac{-2}{5} + i\frac{4}{5}$. The question asks for the complex conjugate $\overline{z}$, which is found by simply changing the sign of the imaginary part. Therefore, $\overline{z} = -\frac{2}{5} - \frac{4}{5}i$. $$ \therefore \text{The conjugate } \overline{z} \text{ is } -\frac{2}{5}-\frac{4}{5}i. $$