Visual SolutionPYQ 2024Tricky

Question

The locus of

z

such that

\arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{2}

is:

(A)A straight line

(B)A circle passing through

1

and

-1

(C)The imaginary axis

(D)The real axis

Solution Path

Angle subtended

= \pi/2

implies semicircle with diameter

[-1, 1]

. Locus is a circle through

1

and

-1

01Question Setup

1/4

Find the locus of

z

satisfying

\arg\!\left(\frac{z-1}{z+1}\right) = \frac{\pi}{2}

\text{Locus of } z = \;?

02Geometric Meaning

2/4

\arg\!\left(\frac{z-1}{z+1}\right) = \frac{\pi}{2}

means the angle at

z

subtended by the segment from

-1

1

\frac{\pi}{2}

\angle(-1, z, 1) = \frac{\pi}{2}

03Semicircle LocusKEY INSIGHT

3/4

By the angle in a semicircle theorem,

z

lies on a semicircle with diameter from

-1

1

. Locus:

x^2 + y^2 = 1

y > 0

x^2 + y^2 = 1, \; y > 0

04Final Answer

4/4

A circle passing through

1

and

-1

\boxed{\text{A circle passing through } 1 \text{ and } -1}

Concepts from this question2 concepts unlocked

★

STANDARD

Conjugate pairs are reflections across the real axis

z = x + iy \implies \bar{z} = x - iy

Recognizing conjugate pairs halves the computation - products simplify dramatically

Locus problemsGeometric interpretationSymmetric root problems

STANDARD

When |z| = 1, the conjugate equals the reciprocal: z̅ = 1/z

|z| = 1 \implies \bar{z} = \frac{1}{z} = e^{-i\theta}

Converts division/conjugate operations into simple exponent changes on the unit circle

Locus problemsModulus equationsProduct simplification

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