Question

If $z_1$ and $z_2$ are complex numbers such that $\mathrm{Re}\left(z_1\right)=\left|z_1-1\right|$, $\mathrm{Re}\left(z_2\right)=\left|z_2-1\right|$, and $\arg \left(z_1-z_2\right)=\frac{\mathrm{\pi }}{6}$, then $\mathrm{Im}\left(z_1+z_2\right)$ is equal to:

• A. $2\sqrt{3}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A

$2\sqrt{3}$

Explanation for the correct option.

Step 1. Assume two complex numbers.

Let $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$ be two complex numbers.

The real part of $z_1$ is $\mathrm{Re}\left(z_1\right)=x_1$ and the real part of $z_2$ is $\mathrm{Re}\left(z_2\right)=x_2$.

Step 2. Form the equation using the given information $\mathrm{Re}\left(z_1\right)=\left|z_1-1\right|$.

It is given that $\mathrm{Re}\left(z_1\right)=\left|z_1-1\right|$, so substitute $\mathrm{Re}\left(z_1\right)=x_1$ and $z_1=x_1+i{y}_1$ .

$$\begin{gathered} \mathrm{Re}\left(z_1\right)=\left|z_1-1\right| \\ \Rightarrow x_1=\left|x_1+i{y}_1-1\right| \\ \Rightarrow x_1=\left|\left(x_1-1\right)+i{y}_1\right| \\ \Rightarrow x_1=\sqrt{{\left(x_1-1\right)}^2+{y_1}^2} \end{gathered}$$

Now, square both sides.

$$\begin{gathered} {x_1}^2={\left(\sqrt{{\left(x_1-1\right)}^2+{y_1}^2}\right)}^2 \\ \Rightarrow {x_1}^2={x_1}^2-2x_1+1+{y_1}^2 \\ \Rightarrow -{y_1}^2=-2x_1+1 \\ {\Rightarrow y_1}^2=2x_1-1...\left(1\right) \end{gathered}$$

Step 3. Form the equation using the given information $\mathrm{Re}\left(z_2\right)=\left|z_2-1\right|$.

It is given that $\mathrm{Re}\left(z_2\right)=\left|z_2-1\right|$, so substitute $\mathrm{Re}\left(z_2\right)=x_2$ and $z_2=x_2+i{y}_2$ .

$$\begin{gathered} \mathrm{Re}\left(z_2\right)=\left|z_2-1\right| \\ \Rightarrow x_2=\left|x_2+i{y}_2-1\right| \\ \Rightarrow x_2=\left|\left(x_2-1\right)+i{y}_2\right| \\ \Rightarrow x_2=\sqrt{{\left(x_2-1\right)}^2+{y_2}^2} \end{gathered}$$

Now, square both sides.

$$\begin{gathered} {x_2}^2={\left(\sqrt{{\left(x_2-1\right)}^2+{y_2}^2}\right)}^2 \\ \Rightarrow {x_2}^2={x_2}^2-2x_2+1+{y_2}^2 \\ \Rightarrow -{y_2}^2=-2x_2+1 \\ {\Rightarrow y_2}^2=2x_2-1...\left(2\right) \end{gathered}$$

Step 4. Form the equation using the given information $\arg \left(z_1-z_2\right)=\frac{\mathrm{\pi }}{6}$.

In the given equation $\arg \left(z_1-z_2\right)=\frac{\mathrm{\pi }}{6}$ substitute $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$.

$$\begin{gathered} \arg \left(x_1+i{y}_1-x_2-i{y}_2\right)=\frac{\mathrm{\pi }}{6} \\ \Rightarrow \arg \left(x_1-x_2+i(y_1-y_2)\right)=\frac{\mathrm{\pi }}{6} \\ \Rightarrow {\tan }^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\frac{\mathrm{\pi }}{6}\left[\arg \left(x+iy\right)={\tan }^{-1}\left(\frac{y}{x}\right)\right] \\ \Rightarrow \frac{y_1-y_2}{x_1-x_2}=\tan \frac{\mathrm{\pi }}{6} \\ \Rightarrow \frac{y_1-y_2}{x_1-x_2}=\frac{1}{\sqrt{3}}...\left(3\right) \end{gathered}$$

Step 5. Find the value of $\mathrm{Im}\left(z_1+z_2\right)$.

Subtract equation $2$ from equation $1$.

$$\begin{gathered} {y_1}^2-{y_2}^2=2x_1-1-2x_2+1 \\ \Rightarrow \left(y_1-y_2\right)\left(y_1+y_2\right)=2\left(x_1-x_2\right)\left[a^2-b^2=(a-b)(a+b)\right] \\ \Rightarrow y_1+y_2=2\frac{x_1-x_2}{y_1-y_2} \\ \Rightarrow y_1+y_2=2\frac{1}{{\displaystyle \frac{y_1-y_2}{x_1-x_2}}} \\ \Rightarrow y_1+y_2=2\frac{1}{{\displaystyle \frac{1}{\sqrt{3}}}}\left[\mathrm{Using}\mathrm{equation}3\right] \\ \Rightarrow y_1+y_2=2\sqrt{3} \end{gathered}$$

Now, $\mathrm{Im}\left(z_1+z_2\right)$ is given as

$\begin{array}{rcl}\mathrm{Im}\left(z_1+z_2\right)& =& \mathrm{Im}\left(x_1+i{y}_1+x_2+i{y}_2\right)\\ & =& \mathrm{Im}\left(x_1+x_2+i\left(y_1+y_2\right)\right)\\ & =& y_1+y_2\left[\mathrm{Im}\left(x+iy\right)=y\right]\\ & =& 2\sqrt{3}\left[y_1+y_2=2\sqrt{3}\right]\end{array}$

So, the value of $\mathrm{Im}\left(z_1+z_2\right)$ is $2\sqrt{3}$.

Hence, the correct option is A.