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Question

Mathematics Question on Complex Numbers and Quadratic Equations

Let z1z_1 and z2z_2 be any two non-zero complex numbers such that 3z1=4z23|z_1| = 4 |z_2|. If z=3z12z2+2z23z1z = \frac{3z_{1}}{2z_{2}} + \frac{2z_{2}}{3z_{1}} then :

A

z=12172\left|z\right| = \frac{1}{2} \sqrt{\frac{17}{2}}

B

Re(z)=0Re(z) = 0

C

z=52\left|z\right| = \sqrt{\frac{5}{2}}

D

Im(z)=0Im(z) = 0

Answer

Im(z)=0Im(z) = 0

Explanation

Solution

3z1=4z23\left|z_{1}\right| = 4 \left|z_{2}\right|
z1z2=43\Rightarrow \frac{\left|z_{1}\right|}{\left|z_{2}\right|} = \frac{4}{3}
3z12z2=2\Rightarrow \frac{\left|3z_{1}\right|}{\left|2z_{2}\right|} = 2
Let 3z12z2=a=2cosθ+2isinθ\frac{3z_{1}}{2z_{2}} = a = 2 \cos \theta + 2 i \sin\theta
z=3z12z2+2z23z1=a+1az = \frac{3z_{1}}{2z_{2}} + \frac{2z_{2}}{3z_{1}} = a+ \frac{1}{a}
=52cosθ+32isinθ= \frac{5}{2} \cos\theta + \frac{3}{2} i \sin \theta
Now all options are incorrect .
There is a misprint in the problem actual
problem should be :
"Let z1z_1 and z2z_2 be any non-zero complex number such that 3z1=2z23|z_1| = 2|z_2|.
If z=3z12z2+2z23z1z = \frac{3z_{1}}{2z_{2}} + \frac{2z_{2}}{3z_{1}} , then "
Given
3z1=2z23 |z_1| = 2 |z_2|
Now 3z12z2=1\left|\frac{3z_{1}}{2z_{2}}\right| = 1
Let 3z12z2=a=cosθ+isinθ\frac{3z_{1}}{2z_{2} } = a = \cos\theta + i \sin\theta
z=3z12z2+2z23z1z = \frac{3z_{1} }{2z_{2}} + \frac{2z_{2}}{3z_{1}}
=a+1a=2cosθ= a + \frac{1}{a} = 2 \cos\theta
Im(z)=0\therefore Im(z) = 0