If for z = + i, |z + 2| = z + 4(1 + i), then + and are the roots of the equation:

Given: z = + i and |z+2| = z + 4(1+i). Find: The equation whose roots are + and. Since |z+2| is a real number, the imaginary part of the right-hand side must be zero. z + 4(1+i) = ( + i) + 4 + 4i = ( +4) + i( +4) So, + 4 = 0 = -4 Now equating the real parts, |z+2| = + 4 But z+2 = ( +2) + i = ( +2) - 4i Hence, |z+2| = ( +2)^2 + 16 Therefore, ( +2)^2 + 16 = + 4 Squaring both sides, ( +2)^2 + 16 = ( +4)^2 ^2 + 4 + 20 = ^2 + 8 + 16 4 = 4 = 1 Thus, + = 1 + (-4) = -3 and = 1 (-4) = -4 So the required quadratic equation with roots -3 and -4 is x^2 - ( -3 + -4)x + (-3)(-4) = 0 x^2 + 7x + 12 = 0 Therefore, the correct option is B.

If for z = + i, |z + 2| = z + 4(1 + i), then + and are the roots… — Solution | JEE Mathematics

MCQMediumJEE 2023Complex Numbers Basics

If for $z = \alpha + i\beta$ , $|z + 2| = z + 4(1 + i)$ , then $\alpha + \beta$ and $\alpha\beta$ are the roots of the equation:

A
$x^2 + 3x - 4 = 0$
B
$x^2 + 7x + 12 = 0$
C
$x^2 - 7x + 12 = 0$
D
$x^2 + 2x - 3 = 0$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $z = \alpha + i\beta$ and $|z+2| = z + 4(1+i)$ .

Find: The equation whose roots are $\alpha + \beta$ and $\alpha\beta$ .

Since $|z+2|$ is a real number, the imaginary part of the right-hand side must be zero.

z + 4(1+i) = (\alpha + i\beta) + 4 + 4i = (\alpha+4) + i(\beta+4)

So,

\beta + 4 = 0 \Rightarrow \beta = -4

Now equating the real parts,

|z+2| = \alpha + 4

But

z+2 = (\alpha+2) + i\beta = (\alpha+2) - 4i

Hence,

|z+2| = \sqrt{(\alpha+2)^2 + 16}

Therefore,

\sqrt{(\alpha+2)^2 + 16} = \alpha + 4

Squaring both sides,

(\alpha+2)^2 + 16 = (\alpha+4)^2

\alpha^2 + 4\alpha + 20 = \alpha^2 + 8\alpha + 16

4 = 4\alpha

\alpha = 1

Thus,

\alpha + \beta = 1 + (-4) = -3

and

\alpha\beta = 1 \cdot (-4) = -4

So the required quadratic equation with roots $-3$ and $-4$ is

x^2 - ( -3 + -4 )x + (-3)(-4) = 0

x^2 + 7x + 12 = 0

Therefore, the correct option is B.

Common mistakes

Treating $|z+2|$ as a complex number is incorrect because modulus is always real and non-negative. First force the imaginary part of $z + 4(1+i)$ to be zero.
Using $|z| = \alpha + i\beta$ is wrong. The modulus of $z+2 = (\alpha+2) + i\beta$ must be computed as $\sqrt{(\alpha+2)^2 + \beta^2}$ .
While forming the quadratic equation from roots, students often use the wrong sign convention. If roots are $r_1$ and $r_2$ , then the equation is $x^2 - (r_1+r_2)x + r_1r_2 = 0$ .

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Answer

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