Visual SolutionPYQ 2024Easy

Question

z = 1 + i

, then

z \cdot \bar{z}

equals:

(A)

2i

(B)

0

(C)

2

(D)

1 + i

Solution Path

z \cdot \bar{z} = |z|^2 = 1^2 + 1^2 = 2

01Question Setup

1/4

Find

z \cdot \bar{z}

for

z = 1 + i

z \cdot \bar{z} = \;?

02Key Property

2/4

z \cdot \bar{z} = |z|^2 = a^2 + b^2

z \cdot \bar{z} = |z|^2

03ComputeKEY INSIGHT

3/4

z \cdot \bar{z} = |z|^2 = 1^2 + 1^2 = 2

1 + 1 = 2

04Final Answer

4/4

z \cdot \bar{z} = 2

\boxed{2}

Concepts from this question1 concepts unlocked

★

EASY

|a + bi|^2 = a^2 + b^2 - no square root needed

|z|^2 = z \cdot \bar{z} = x^2 + y^2

Final step in many JEE problems. Students waste time square-rooting when they only need |z|^2

Distance problemsSum/product modulus\alpha^2 + \beta^2 type problems

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