Question 4 - Jee advanced Math 2022 P1 Questions with Solutions

Let \(z\) be a complex number with non-zero imaginary part. If

\(\frac{2+3z+4z^{2}}{2-3z+4z^{2}}\)

is a real number, then the value of \(|z|^{2}\) is _____.

Sol : 

It is given that the fraction \(\frac{2+3z+4z^{2}}{2-3z+4z^{2}}\) is a real number.

Let \(p\) be a real number such that 

\(\frac{2+3z+4z^{2}}{2-3z+4z^{2}} = p\)

\(\implies 2+3z+4z^{2} = p(2-3z+4z^{2})\)

\(\implies 2(1-p) + 3z(1+p) + 4z^{2}(1-p) = 0\)

\(\implies (1-p)(2+4z^{2}) + 3z(1+p) = 0\)

\(\implies (1-p)(2+4z^{2}) = -3z(1+p)\)

\(\implies \frac{2+4z^{2}}{3z} = -\frac{(1+p)}{(1-p)}\)
(We are dividing both sides by \((1-p)\) because \(p \neq 1\). If \(p=1 \implies 2+3z+4z^{2} = 1(2-3z+4z^{2}) \implies z = 0 + 0 i\), which contradicts the given fact that \(z\) has non-zero imaginary part.)

\(\implies \frac{2+4z^{2}}{3z} = \frac{(1+p)}{(p-1)}\)

As \(p\) is a real number, \(\frac{1+p}{p-1}\) is also a real number.
\(\implies \frac{2+4z^{2}}{3z}\) has imaginary part equal to zero.

\(\frac{2+4z^{2}}{3z} = \frac{2}{3z} + \frac{4z}{3}\)

Let \(z=a+bi\)

\(\frac{2+4z^{2}}{3z} = \frac{2(a-bi)}{3(a+bi)(a-bi)} + \frac{4(a+bi)}{3}\)

             \( = \frac{2(a-bi)}{3(a^{2}+b^{2})} + \frac{4(a+bi)}{3}\)

Imaginary part \(= \frac{-2b}{3(a^{2}+b^{2})} + \frac{4b}{3} = 0\)

\(\implies \frac{1}{a^{2}+b^{2}} = 2\)
\(\implies a^{2}+b^{2} = \frac{1}{2}\)

Therefore,
 \(|z|^{2} = (\sqrt{a^{2}+b^{2}})^{2}\)
                    \( = a^{2} + b^{2}\)
                     \(\frac{1}{2} = 0.5\)






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