The set of all α ∈ R , for which w=(1+(1-8 α) z)/(1-z) is a purely imaginary number, for all z ∈ C satisfying |z|=1 and Re z ≠ 1 , is

Correct answer: A. \ 0\ aligned & |z|=1 & Rez ≠ 1 & Suppose z=x+i y x²+y²=1 (i) & Now, w=(1+(1-8 α) z)/(1-z) & w=(1+(1-8 α)(x+i y))/(1-(x+i y)) & w= 1+(1-8 α)(x+i y))((1-x)+iy) 1-(x+i y))((1-x)+iy) & w=([(1+x(1-8 α)(1-x)-(1-8) y^2].)/((1-x)^2+y^2) &+i ([(1+x(1-8 α)) y-(1-8 α) y(1-x)])/((1-x)^2+y^2) aligned If, w is purely imaginary. So, aligned &Re w=([(1+x(1-8 α))(1-α)-(1-8 α) y^2])/((1-x)^2+y^2) &=0 & (1-x)+x(1-8 α)(1-x)=(1-8) y² & (1-x)+x(1-8 α)-x^2(1-8 α)=(1-8 x) y² & (1-x)+x(1-8 α)=1-8 α & [ From (i), x²+y²=1] aligned 1-8 α=1 α=0 α ∈\ 0\

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Mathematics
Chapter: Complex Number

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Complex Number

The set of all α ∈ R , for which w=(1+(1-8 α) z)/(1-z) is a purely...

The set of all

\alpha \in R

, for which

w=\frac{1+(1-8 \alpha) z}{1-z}

is a purely imaginary number, for all

z \in C

satisfying

|z|=1

and

\operatorname{Re} z \neq 1

, is

Options

A $\{0\}$
Ban empty set
C $\left\{0, \frac{1}{4},-\frac{1}{4}\right\}$
Dequal to $R$

Correct answer

A. $\{0\}$

Step-by-step solution

\begin{aligned}&\because|z|=1 \text{ & } \operatorname{Re}z ≠ 1 \\ &\text{ Suppose }z=x+i y \Rightarrow x²+y²=1 \ldots \ldots(\mathrm{i})\\ &\text{ Now, }w=\frac{1+(1-8 \alpha) z}{1-z}\\ &\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y)}{1-(x+i y)}\\ &\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y))((1-\mathrm{x})+\mathrm{iy})}{1-(x+i y))((1-\mathrm{x})+\mathrm{iy})}\\ &\Rightarrow \quad w=\frac{\left[\left(1+x(1-8 \alpha)(1-x)-(1-8) y^2\right]\right.}{(1-x)^2+y^2}\\ &+i \frac{[(1+x(1-8 \alpha)) y-(1-8 \alpha) y(1-x)]}{(1-x)^2+y^2} \end{aligned}

If,

w

is purely imaginary. So,

\begin{aligned}&\operatorname{Re} w=\frac{\left[(1+x(1-8 \alpha))(1-\alpha)-(1-8 \alpha) y^2\right]}{(1-x)^2+y^2}\\ &=0 \\ &\Rightarrow(1-x)+x(1-8 \alpha)(1-x)=(1-8) y² \\ &\Rightarrow(1-x)+x(1-8 \alpha)-x^2(1-8 \alpha)=(1-8 x) y² \\ &\Rightarrow(1-x)+x(1-8 \alpha)=1-8 \alpha\\ &{\left[\text{\text{ From } }(i), x²+y²=1\right]} \end{aligned}

\Rightarrow 1-8 \alpha=1\Rightarrow \alpha=0\therefore \alpha \in\{0\}$

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