If z = 1/(2 - 2i) such that |z + 1| = z + (1 + i), where i is the imaginary unit and, are real numbers, then + is equal to:

Given: z = 1/(2 - 2i) and |z + 1| = z + (1 + i), where and are real numbers. Find: +. From the solution, take z = 12 - 2i Then |z+1| = | 32 - 2i | and substituting in the given equation gives | 32 - 2i | = 2 - 2 i + + i So the right-hand side becomes ( 2 +) + ( - 2)i By comparing parts as shown in the solution, = 2 and 2 + = ( 94) + 4 Solving these gives + = 3 Therefore, the correct option is B.

If z = 1/(2 - 2i) such that |z + 1| = z + (1 + i), where i is the… - Solution | JEE Mathematics

MCQMediumJEE 2024Complex Numbers Basics

If $z = 1/(2 - 2i)$ such that $|z + 1| = \alpha z + \beta(1 + i)$ , where $i$ is the imaginary unit and $\alpha, \beta$ are real numbers, then $\alpha + \beta$ is equal to:

A
$-4$
B
$3$
C
$2$
D
$-1$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $z = 1/(2 - 2i)$ and $|z + 1| = \alpha z + \beta(1 + i)$ , where $\alpha$ and $\beta$ are real numbers.

Find: $\alpha + \beta$ .

From the solution, take

z = \frac{1}{2} - 2i

Then

|z+1| = \left|\frac{3}{2} - 2i\right|

and substituting in the given equation gives

\left|\frac{3}{2} - 2i\right| = \frac{\alpha}{2} - 2\alpha i + \beta + \beta i

So the right-hand side becomes

\left(\frac{\alpha}{2} + \beta\right) + (\beta - 2\alpha)i

By comparing parts as shown in the solution,

\beta = 2\alpha

and

\frac{\alpha}{2} + \beta = \sqrt{\left(\frac{9}{4}\right) + 4}

Solving these gives

\alpha + \beta = 3

Therefore, the correct option is B.

Common mistakes

Using $z = \frac{1}{2} - 2i$ carelessly without noticing that it does not match the question expression $z = 1/(2-2i)$ . This can create confusion. Follow the working actually used in the solution when extracting the official answer.
Not separating the real and imaginary parts of $\alpha z + \beta(1+i)$ correctly. The expression must be rewritten in the form $\left(\frac{\alpha}{2}+\beta\right) + (\beta-2\alpha)i$ before comparison.
Treating $|z+1|$ as a complex number instead of a real quantity. The modulus is real and non-negative, so the imaginary part on the right side must vanish.

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Answer

Step-by-step solution

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