JEE Complex Numbers 1

$\dfrac{1+z+z^2}{1-z+z^2}\in R\\\Rightarrow 1+\dfrac{2z}{1-z+z^2}\in R\\\Rightarrow \dfrac{z}{1-z+z^2}\in R\\\Rightarrow \dfrac{1-z+z^2}{z}\in R\\\Rightarrow z+\dfrac{1}{z}\in R\\\Rightarrow z+\dfrac{1}{z}=\bar{z}+\dfrac{1}{\bar{z}}\\\Rightarrow z-\bar{z}-\left (\dfrac{1}{z}-\dfrac{1}{\bar{z}}\right )=0\\\Rightarrow (z-\bar{z})\left (1-\dfrac{1}{z\bar{z}}\right )=0$

Since $z\in C-R,z\ne \bar{z}$ . Thus, $z\bar{z}=|z|^2=1\Rightarrow |z|=1$

I'm sure that a proof can be generated by saying $a=\bar{a}$

Where $a$ is the given expression.

But in the spirit of JEE, a more expedient approach is to note that $z=i$ satisfies the required relation.

Thus, $|z|=1$ .

put z=i ... the rest is simple !

Let $z=a+ib$ . In this case since $z \in \mathbb{C} - \mathbb{R}$ , we can say $b\neq 0$ $$ Now since $\dfrac{1+z+z^2}{1-z+z^2} \in \mathbb{R}, \implies \dfrac{(1+z+z^2)+(1-z+z^2)}{(1+z+z^2)-(1-z+z^2)} \in \mathbb{R}$ $$ This means $\dfrac{1+z^2}{z} \in \mathbb{R}$ $$ Now if we replace $z$ by $a+ib$ , multiply the denominator by $\bar{z}$ ,and then simply setting the imaginary part of the resultant expression equal to $0$ , we get the equation $$ $b=b^3+a^2b \implies \sqrt{a^2+b^2} = |z| = 1$ $($ Since $b\neq 0$ $)$ $$ This involves some tedious calculations. I'm sure there exists a nice solution for this!

• Rearranging the expression as $\frac{(1 + z^{2}) + z}{(1 + z^{2}) - z}$ , we can easily see $z = i$ as a solution, since $1 + i^{2} = 0$ , leaving $\frac{i}{-i} = -1$ .
• Since $|z| = \sqrt{a^{2} + b^{2}}$ , we have $|z|= |i| = 1 + 0 = 1$ .

simply multiply both num-tor and den- tor by-

                     (z^(2) + 1)+(z)

and simplify with the same bracketed values further you get a pallindrome equation in the complex no 'z' ,in the numerator of degree 4, to which you divide both num-tor and den-tor by

                            ((z)^2)

when you recollect the terms together , it is obvious to get the generalisation that for the whole part to be real
(1/z) need to be the conjugate of z. hence consequently we reach to the answer 1.