Question

Let ${{z}_{1}}$ and ${{z}_{2}}$ be any two non-zero complex numbers such that $3\left| {{z}_{1}} \right|=4\left| {{z}_{2}} \right|$. If $z=\dfrac{3{{z}_{1}}}{2{{z}_{2}}}+\dfrac{2{{z}_{2}}}{3{{z}_{1}}}$ then?
A. $\left| z \right|=\sqrt{\dfrac{17}{2}}$
B. $\operatorname{Re}(z)=0$
C. $\left| z \right|=\sqrt{\dfrac{5}{2}}$
D. $\operatorname{Im}(z)=0$

Answer 1

Hint: In order to solve this question we have to manipulate the given equation $3\left| {{z}_{1}} \right|=4\left| {{z}_{2}} \right|$. From this equation get the value of $\left| \dfrac{3{{z}_{1}}}{2{{z}_{2}}} \right|=2$and after getting this as we know that from the Euler’s representation of a complex number, any complex number $a$ can be represented as $\left| a \right|{{e}^{i\theta }}$ or $\left| a \right|\cos \theta +i\left| a \right|\sin \theta $ . Suppose that the $x=\dfrac{3{{z}_{1}}}{2{{z}_{2}}}$ then $x$can be written as $\left| \dfrac{3{{z}_{1}}}{2{{z}_{2}}} \right|{{e}^{i\theta }}$i.e. $2{{e}^{i\theta }}$. Now $z$ can be written as $x+\dfrac{1}{x}$ and then evaluate the value of $z$ by putting $x=2{{e}^{i\theta }}$ and then get the answer.

Complete step-by-step answer:
It is given that $3\left| {{z}_{1}} \right|=4\left| {{z}_{2}} \right|$
Taking $2\left| {{z}_{2}} \right|$ in the LHS, we get
$\Rightarrow \dfrac{3\left| {{z}_{1}} \right|}{2\left| {{z}_{2}} \right|}=2$
We can also write the above equation as $\left| \dfrac{3{{z}_{1}}}{2{{z}_{2}}} \right|=2$
Let us assume that $x=\dfrac{3{{z}_{1}}}{2{{z}_{2}}}$.
Now as we know that from the Euler’s representation of a complex number, any complex number $a$ can be represented as $\left| a \right|{{e}^{i\theta }}$ or $\left| a \right|\cos \theta +i\left| a \right|\sin \theta $ where $\theta $ the argument of $a$.
Hence, $x$can also be represented as $x=\left| \dfrac{3{{z}_{1}}}{2{{z}_{2}}} \right|{{e}^{i\theta }}=2{{e}^{i\theta }}$
Similarly $\dfrac{1}{x}$ can be represented as $\dfrac{1}{x}=\dfrac{1}{2{{e}^{i\theta }}}=\dfrac{{{e}^{-i\theta }}}{2}$
Now it is given that $z=\dfrac{3{{z}_{1}}}{2{{z}_{2}}}+\dfrac{2{{z}_{2}}}{3{{z}_{1}}}$, putting the value $x=\dfrac{3{{z}_{1}}}{2{{z}_{2}}}$ in this equation, we get
$\Rightarrow z=x+\dfrac{1}{x}$
Substituting the value of $x=2{{e}^{i\theta }}$ and $\dfrac{1}{x}=\dfrac{{{e}^{-i\theta }}}{2}$ in the above equation, we get
$\Rightarrow z=2{{e}^{i\theta }}+\dfrac{{{e}^{-i\theta }}}{2}$
Now we know that ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ and ${{e}^{-i\theta }}=\cos \theta -i\sin \theta $, substituting both these in the above equation, we get
$\Rightarrow z=2\left( \cos \theta +i\sin \theta \right)+\dfrac{1}{2}\left( \cos \theta -i\sin \theta \right)$
$\Rightarrow z=\dfrac{5}{2}\cos \theta +i\dfrac{3}{2}\sin \theta $
Hence, the value of $z$ is equal to $\dfrac{5}{2}\cos \theta +i\dfrac{3}{2}\sin \theta $.
So, $\operatorname{Re}\left( z \right)=\dfrac{5}{2}\cos \theta \ne 0$ and $\operatorname{Im}\left( z \right)=\dfrac{3}{2}\sin \theta \ne 0$
Now, $\left| z \right|=\sqrt{{{\left( \dfrac{5}{2} \right)}^{2}}+{{\left( \dfrac{3}{2} \right)}^{2}}}=\sqrt{\dfrac{25}{4}+\dfrac{9}{4}}=\sqrt{\dfrac{34}{4}}=\sqrt{\dfrac{17}{2}}$

So, the correct answer is “Option A”.

Note: This question is a typical example of Euler representation of a complex number. Students should note two things, first one is the given equation which is $3\left| {{z}_{1}} \right|=4\left| {{z}_{2}} \right|$, we can rearrange it to get the modulus of the complex number $\dfrac{3{{z}_{1}}}{2{{z}_{2}}}$. Second thing to notice that $z$ is of the form $x+\dfrac{1}{x}$ , so if we represent $x=\left| x \right|{{e}^{i\theta }}$ then $z$ can be easily found as \[z=x+\dfrac{1}{x}=\left| x \right|{{e}^{i\theta }}+\dfrac{{{e}^{-i\theta }}}{\left| x \right|}\]. Hence, as the modulus of $x$ is known so we represented it like that rather than assuming $x=a+ib$ and hence made the calculation a lot easier.