Question

Number of complex number z such that $\left| z \right|=1$ and $\left| \dfrac{z}{\overline{z}}+\dfrac{\overline{z}}{z} \right|=1$ is?
(a) 4
(b) 6
(c) 8
(d) More than 8

Answer 1

Assume $z=x+iy$ and write its conjugate given as $\overline{z}=x-iy$. Now, use the formula $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ to form first relation between x and y considering the relation $\left| z \right|=1$. Simplify the relation $\left| \dfrac{z}{\overline{z}}+\dfrac{\overline{z}}{z} \right|=1$ to form a second relation between x and y. Solve the two relations to form sets of values of x and y and count the number of solutions. Use the identity of modulus given as if $\left| x \right|=a$ then $x=\pm a$.

Complete step by step answer:
Here we have been provided with the complex number z and two relations $\left| z \right|=1$ and $\left| \dfrac{z}{\overline{z}}+\dfrac{\overline{z}}{z} \right|=1$. We are asked to find the number of complex numbers satisfying the given relations.
Now, let us assume the complex number as $z=x+iy$, so its conjugate will be given as $\overline{z}=x-iy$. We know that modulus of a complex number is given as $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$, so considering the relation $\left| z \right|=1$ we get,
$\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}}=1$
On squaring both the sides we get,
$\Rightarrow {{x}^{2}}+{{y}^{2}}=1$ ……….. (1)
Now, let us simplify the relation $\left| \dfrac{z}{\overline{z}}+\dfrac{\overline{z}}{z} \right|=1$, so we have,

& \Rightarrow \left| \left( \dfrac{x+iy}{x-iy} \right)+\left( \dfrac{x-iy}{x+iy} \right) \right|=1 \\\ & \Rightarrow \left| \dfrac{{{\left( x+iy \right)}^{2}}+{{\left( x-iy \right)}^{2}}}{\left( x-iy \right)\left( x+iy \right)} \right|=1 \\\ \end{aligned}$$ Using the three basic algebraic identities given as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, substituting the value ${{i}^{2}}=-1$ we get, $$\begin{aligned} & \Rightarrow \left| \dfrac{\left( {{x}^{2}}-{{y}^{2}}+2ixy \right)+\left( {{x}^{2}}-{{y}^{2}}-2ixy \right)}{{{x}^{2}}+{{y}^{2}}} \right|=1 \\\ & \Rightarrow \left| \dfrac{2\left( {{x}^{2}}-{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}} \right|=1 \\\ \end{aligned}$$ Substituting the value ${{x}^{2}}+{{y}^{2}}=1$ in the above relation using relation (1) we get, $$\begin{aligned} & \Rightarrow \left| 2\left( {{x}^{2}}-{{y}^{2}} \right) \right|=1 \\\ & \Rightarrow \left| {{x}^{2}}-{{y}^{2}} \right|=\dfrac{1}{2} \\\ \end{aligned}$$ Since we don’t know whether the expression inside the modulus sign is greater than or less than 0, so using the identity of modulus function given as if $\left| x \right|=a$ then $x=\pm a$ we get, $$\Rightarrow {{x}^{2}}-{{y}^{2}}=\pm \dfrac{1}{2}$$ (i) Considering the expression $${{x}^{2}}-{{y}^{2}}=\dfrac{1}{2}$$ we get on substituting the value of ${{y}^{2}}$ from equation (1), $$\begin{aligned} & \Rightarrow {{x}^{2}}-\left( 1-{{x}^{2}} \right)=\dfrac{1}{2} \\\ & \Rightarrow 2{{x}^{2}}=\dfrac{3}{2} \\\ & \Rightarrow {{x}^{2}}=\dfrac{3}{4} \\\ \end{aligned}$$ Taking square root both the sides we get, $$\Rightarrow x=\pm \dfrac{\sqrt{3}}{2}$$ Here we have obtained two values of x so for each value of x there will be two values of y, therefore there will be 4 different solutions for this case. (ii) Considering the expression $${{x}^{2}}-{{y}^{2}}=-\dfrac{1}{2}$$ we get on substituting the value of ${{y}^{2}}$ from equation (1), $$\begin{aligned} & \Rightarrow {{x}^{2}}-\left( 1-{{x}^{2}} \right)=-\dfrac{1}{2} \\\ & \Rightarrow 2{{x}^{2}}=\dfrac{1}{2} \\\ & \Rightarrow {{x}^{2}}=\dfrac{1}{4} \\\ \end{aligned}$$ Taking square root both the sides we get, $$\Rightarrow x=\pm \dfrac{1}{2}$$ Here also we have obtained two values of x so for each value of x there will be again two values of y, therefore there will be 4 different solutions for this case also. So for the two different cases we have 4 solutions for each case, so the total number of solutions would be 8. However, we haven’t been provided with the information about the argument of the complex number which is given as $\arg \left( z \right)={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$. The argument denotes the angle that is formed by the complex number with the x axis in the Argand plane. So for a particular set of values of x and y there can be an infinite number of angles in the range of real numbers because we know that the trigonometric functions are periodic in nature. **So, the correct answer is “Option d”.** **Note:** Note that there is a different method also to solve this question. In that we will write the Euler’s form of the complex number given as $z=r{{e}^{i\theta }}$, where r is given as $r=\left| z \right|$. In this case we have to focus only on the values of $\theta $ because the value r = 1 is already given. In this case you will get the relation $\left| \cos 2\theta \right|=\dfrac{1}{2}$ which will have an infinite number of solutions because the range of $\theta $ isn’t provided to us.