QJEEM19Maae0a: JEE Main 2019 Mathematics - Class 11 JEEM19Maae0a Solution: Let z 1 and z 2 be any two non-zero complex numbers such…

Question

Verified Solution • 2025–26 Session moderate

Let z₁ and z₂ be any two non-zero complex numbers such that

3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.

If

z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}

then :

Accepted Answer

Given,

3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|

\Rightarrow

{{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}

\Rightarrow

{{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2

As we know, for any compled number

{{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}

(cos

\theta

+ i sin

\theta

)

= 2(cos

\theta

+ i sin

\theta

)

\therefore

{{2{z_2}} \over {3{z_1}}}

=

{1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}

=

{1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}

=

{\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}

Now, given

z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}

= 2(cos

\theta

+ i sin

\theta

) +

{\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}

=

{{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }

So, |z| =

\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta }

=

{1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }

z is neither purely real nor purely imaginary and |z| depends on

\theta

.

Step-by-Step Verified Solution