JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

z=x+i y

and

a

is a real number such that

|z-a i|=|z+a i|

, then locus of

z

View answer and explanation

Correct option: A

X

-axis

Option A (

X

-axis) is correct because the condition |z-ai| = |z+ai| implies that z is equidistant from ai and -ai, which means z lies on the x-axis (y=0).

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