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# If ${{z}_{1}}\text{ and }{{z}_{2}}$are two complex numbers such that ${{z}_{1}}+i\overline{{{z}_{2}}}=0$ and $\arg \left( \overline{{{z}_{1}}}{{z}_{2}} \right)=\dfrac{\pi }{3}$, then $\arg (\overline{{{z}_{1}}})$ is equal to a)$\dfrac{\pi }{3}$b)$\pi$c)$\dfrac{\pi }{2}$d)$\dfrac{5\pi }{12}$e)$\dfrac{5\pi }{6}$

Hint: We need to use the fact that when two complex numbers are multiplied, their argument gets added and then express I in its polar form.

It has been given that ${{z}_{1}}+i\overline{{{z}_{2}}}=0$. However, as we have to find $\arg (\overline{{{z}_{1}}})$, it will be useful to convert the equation in terms of $\overline{{{z}_{1}}}$. For this we take the complex conjugate of the whole expression as:

\begin{align} & {{z}_{1}}+i\overline{{{z}_{2}}}=0\Rightarrow \overline{{{z}_{1}}+i\overline{{{z}_{2}}}}=\overline{0}=0 \\ & \Rightarrow \overline{{{z}_{1}}}-i{{z}_{2}}=0\text{ }............(1.1) \\ \end{align}

Notice that there is a negative sign as while taking the conjugate I becomes –i.

From equation (1.1) we find that

$\overline{{{z}_{1}}}=i{{z}_{2}}\text{ }\Rightarrow {{\text{z}}_{2}}=\dfrac{1}{i}\overline{{{z}_{1}}}=\dfrac{i}{{{i}^{2}}}\overline{{{z}_{1}}}=\dfrac{i}{-1}\overline{{{z}_{1}}}=-i\overline{{{z}_{1}}}\text{ }..........(1.2)$

Now a complex number z can be represented by $z=r{{e}^{i\theta }}$ which is known as the polar form. The argument of z is the angle $\theta$ in the polar form. Euler’s formula states that ${{e}^{i\theta }}=r\left( \cos (\theta )+i\sin (\theta ) \right)$.

As -i can be represented as $-i={{e}^{i\left( \dfrac{-\pi }{2} \right)}}$, we get $\arg (-i)=\dfrac{-\pi }{2}\text{ }.............\text{(1}\text{.3)}$.

Now we use the fact that when two complex numbers are multiplied their argument gets added. Thus, from the second relation we obtain

$\arg \left( \overline{{{z}_{1}}}{{z}_{2}} \right)=\dfrac{\pi }{3}\Rightarrow \arg \left( \overline{{{z}_{1}}} \right)+\arg \left( {{z}_{2}} \right)=\dfrac{\pi }{3}\text{ }.........\text{(1}\text{.4)}$

Using (1.2), we replace the value of z2 to obtain

$\arg \left( \overline{{{z}_{1}}} \right)+\arg \left( {{z}_{2}} \right)=\arg \left( \overline{{{z}_{1}}} \right)+\arg \left( -i\overline{{{z}_{1}}} \right)=\arg \left( \overline{{{z}_{1}}} \right)+\arg \left( -i \right)+\arg \left( \overline{{{z}_{1}}} \right)=2\arg \left( \overline{{{z}_{1}}} \right)+\left( \dfrac{-\pi }{2} \right)\text{ }......\text{(from equation(1}\text{.3))}$

Using this in equation (1.4), we get

\begin{align} & \arg \left( \overline{{{z}_{1}}} \right)+\arg \left( {{z}_{2}} \right)=\dfrac{\pi }{3}\text{ }\Rightarrow \text{2}\arg \left( \overline{{{z}_{1}}} \right)-\dfrac{\pi }{2}=\dfrac{\pi }{3} \\ & \Rightarrow \arg \left( \overline{{{z}_{1}}} \right)=\dfrac{\dfrac{\pi }{2}+\dfrac{\pi }{3}}{2}=\dfrac{5\pi }{12} \\ \end{align}

Thus, as this answer matches with the value in option (d), option (d) is the correct answer.

Note: These kinds of questions are most easily solved by taking the polar form instead of using the normal form of a complex number z=x+iy. However, one can still solve it using this form but the steps will be complicated and we have to derive the angles using the inverse trigonometric functions which are difficult to handle.