Complex Numbers JEE Main Mathematics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: a

Explanation: the X-axis ​$\Rightarrow y^{2} - 10 y + 25 = y^{2} + 10 y + 25$​​ ​$\Rightarrow \left(\right. y - 5 \left.\right)^{2} = \left(\right. y + 5 \left.\right)^{2}$​​ ​$\Rightarrow y^{2} - 10 y + 25 = y^{2} + 10 y + 25$​​ ​$\Rightarrow 20 y = 0$​​ ​$\Rightarrow y = 0$​​ ​∴  _z_ lies on _x_-axis.​

Correct Answer: b

Explanation: Three collinear points $\begin{matrix}W e h a v e , \\ 2 z_{1} 3 z_{2} + z_{3} = 0 \\ z_{2} = \frac{2 z_{1} + z_{3}}{2 + 1} \\ z_{2} d i v i d e s t h e l i n e s e g m e n t j o i n i n g z_{1} a n d z_{3} i n t h e r a t i o 1 : 2 \\ z_{1} , z_{2} , z_{3} a r e c o l l i n e a r.\end{matrix}$

Correct Answer: d

Explanation: None of these ​$\left(x + i y\right)^{\frac{1}{3}} = a + i b$ ​

Correct Answer: c

Explanation: $3 z_{0}^{2}$ $\begin{matrix}S i n c e z_{1} , z_{2} , z_{3} a r e v e r t i c e s o f a n e q u i l a t e r a l t r i a n g l e \\ z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} \\ \left(z_{1} + z_{2} + z_{3}\right)^{2} = 3 \left(z_{1}^{2} + z_{2}^{2} + z_{3}^{2}\right) \\ \left(3 z_{0}\right)^{2} = 3 \left(z_{1}^{2} + z_{2}^{2} + z_{3}^{2}\right) \\ \left[ \frac{z_{1} + z_{2} + z_{3}}{3} = z_{0}\right] \\ z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}\end{matrix}$

Correct Answer: (c)

Explanation: Given $n_1, n_2, n_3$ are roots of the equation $ \begin{array}{r} E_1: x^3+x^2+l x+n=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \therefore \text { Sum of roots } x_1+x_2+x_3=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $ Now, $E_2: x^3+a x^2+b x+c=0$ is reciprocal equation of class one $\Rightarrow c=1$ and $a=b$ $ \therefore \quad E_2: x^3+a x^2+a x+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$ Given, Eq. (iii) have roots $ \begin{aligned} & \frac{x_1-1}{2}, \frac{x_2-1}{2}, \frac{x_3-1}{2} \\ \Rightarrow & \frac{x_1-1}{2}+\frac{x_2-1}{2}+\frac{x_3-1}{2}=-a \\ \Rightarrow & \frac{x_1+x_2+x_3}{2}-\frac{3}{2}=-a \end{aligned} $ From Eq. (ii), $ \begin{aligned} & \quad x_1+x_2+x_3=-1 \\ & \therefore \quad \frac{-1}{2}-\frac{3}{2}=-a \Rightarrow a=2 \\ & \therefore \text { Eq. }(\text { iii }), x^3+2 x^2+2 x+1=0 \\ & (x+1)\left(x^2+x+1\right)=0 \quad \therefore \quad x=-1, \\ & x^2+x+1=0 \end{aligned} $ $ \begin{aligned} & \Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2} \\ & \therefore \text { Given, } \frac{x_1-1}{2}=-1 \Rightarrow x_1=-1 \\ & \quad \frac{x_2-1}{2}=\frac{-1+\sqrt{3} i}{2} \\ & \Rightarrow \quad x_2=\sqrt{3} i \\ & \quad \frac{x_3-1}{2}=\frac{-1-\sqrt{3} i}{2} \\ & \Rightarrow \quad x_3=-\sqrt{3} i \end{aligned} $ ∴ Roots of these two equation excluding common roots are $ \sqrt{3} i,-\sqrt{3} i, \frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2} $