If the least and the largest real values of $\alpha$, for which the equation $z+\alpha|z-1|+2 i=0(z \in C$ and $i =\sqrt{-1}$ ) has a solution, are $p$ and $q$ respectively, then $4\left(p^{2}+q^{2}\right)$ is equal to

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Complex Numbers

If the least and the largest real values of

α

, for which the equation

z + α ∣ z - 1∣ + 2 i = 0 (z \in C

and

i = - 1

) has a solution, are

p

and

q

respectively, then

4 (p^{2} + q^{2})

is equal to

Answer: 10

Let

z = x + i y

. Then

∣ z - 1∣ = (x - 1)^{2} + y^{2}

.

The equation becomes:

x + i y + α (x - 1)^{2} + y^{2} + 2 i = 0

Separating real and imaginary parts:

x + α (x - 1)^{2} + y^{2} = 0

y + 2 = 0 \Rightarrow y = - 2

Substituting

y = - 2

into the real part:

x + α (x - 1)^{2} + 4 = 0

α = - \frac{x}{( x - 1 ) ^{2} + 4}

For

α

to be real, we need to find the range. Let

f (x) = - \frac{x}{( x - 1 ) ^{2} + 4}

.

To find extrema, differentiate and set to zero. After calculation, the minimum value is

p = - 2

and maximum is

q = 1

.

Thus:

4 (p^{2} + q^{2}) = 4 (4 + 1) = 20

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer