Verified Solution • 2025–26 Session moderate

QJEEM24Ma9476: JEE Main 2024 Mathematics - Class 11 JEEM24Ma9476 Solution: If is such that and , then is equal to

If z=12−2iz=\frac{1}{2}-2 i is such that ∣z+1∣=αz+β(1+i),i=−1|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1} and α,β∈R\alpha, \beta \in \mathbb{R} , then α+β\alpha+\beta is equal to

Multiple Choice Options

  • A 2
  • B −- 4
  • C 3
  • D −- 1

Step-by-Step Verified Solution

To begin with, let's analyze the given equation:

∣z+1∣=αz+β(1+i)|z+1|=\alpha z+\beta(1+i)

First, we compute the modulus of the left side:

Let's take the given value of z,

z=12−2iz = \frac{1}{2} - 2i

Now, we find z+1z + 1 :

z+1=(12−2i)+1=32−2iz + 1 = \left(\frac{1}{2} - 2i\right) + 1 = \frac{3}{2} - 2i

Then, the modulus of z+1z + 1 is calculated as follows:

∣z+1∣=∣32−2i∣=(32)2+(−2)2|z + 1| = \left|\frac{3}{2} - 2i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2}

∣z+1∣=94+4|z + 1| = \sqrt{\frac{9}{4} + 4}

∣z+1∣=94+164|z + 1| = \sqrt{\frac{9}{4} + \frac{16}{4}}

∣z+1∣=254|z + 1| = \sqrt{\frac{25}{4}}

∣z+1∣=52|z + 1| = \frac{5}{2}

Now we need to equate the modulus to the right-hand side of the equation and solve for α\alpha and β\beta . Let's rewrite the equation:

52=αz+β(1+i)\frac{5}{2} = \alpha z + \beta(1 + i)

Substitute zz with its value:

52=α(12−2i)+β(1+i)\frac{5}{2} = \alpha \left(\frac{1}{2} - 2i\right) + \beta(1+i)

Rewrite the equation separating real and imaginary parts:

52=α(12)−2αi+β+βi\frac{5}{2} = \alpha \left(\frac{1}{2}\right) - 2\alpha i + \beta + \beta i

52=(α12+β)+(−2α+β)i\frac{5}{2} = \left(\alpha \frac{1}{2} + \beta\right) + \left(-2\alpha + \beta\right)i

For the above equality to hold, both real and imaginary parts must be equal.

Equating real parts:

α12+β=52\alpha \frac{1}{2} + \beta = \frac{5}{2}

Equating imaginary parts:

−2α+β=0-2\alpha + \beta = 0

We now have a system of two linear equations:

1) α2+β=52\frac{\alpha}{2} + \beta = \frac{5}{2}

2) −2α+β=0-2\alpha + \beta = 0

Let's solve the system by isolating β\beta from the second equation and then substituting it into the first one:

β=2α\beta = 2\alpha

Now substitute β\beta in the first equation:

α2+2α=52\frac{\alpha}{2} + 2\alpha = \frac{5}{2}

α(12+2)=52\alpha \left(\frac{1}{2} + 2\right) = \frac{5}{2}

α(12+42)=52\alpha \left(\frac{1}{2} + \frac{4}{2}\right) = \frac{5}{2}

α(52)=52\alpha \left(\frac{5}{2}\right) = \frac{5}{2}

To find the value of α\alpha , we divide both sides by 52\frac{5}{2} :

α=1\alpha = 1

Now, we use the value of α\alpha to find β\beta :

β=2α\beta = 2\alpha

β=2⋅1\beta = 2 \cdot 1

β=2\beta = 2

Finally, we add both α\alpha and β\beta to find α+β\alpha + \beta :

α+β=1+2=3\alpha + \beta = 1 + 2 = 3

The value of α+β\alpha + \beta is 3.

So, the correct answer is Option C) 3.

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