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Complex Numbers question

2023 · 25 Jan · Shift 2 · Q31

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Complex Numbers
2023 · 25 Jan · Shift 2 · Q31

Complex Numbers question

2023 · 25 Jan · Shift 2 · Q31

JEE MainMathematicsComplex NumbersMCQ+4 / −1

Let

z

be a complex number such that

\left| {{{z - 2i} \over {z + i}}} \right| = 2,z e - i

. Then

z

lies on the circle of radius 2 and centre :

A
(0, $-$ 2)
B
(0, 0)
C
(0, 2)
D
(2, 0)

View written solutionFree

Correct answer: A

Interpret the condition

The printed question appears to have a typo. It is intended as:

$\left|\frac{z-2i}{z+i}\right|=2, \quad z\ne -i$

We must find the locus of $z$ and identify the centre of the resulting circle of radius $2$ .

Let $z=x+iy$

Write $z=x+iy$ where $x,y\in\mathbb R$ .

Then $z-2i=x+i(y-2), \qquad z+i=x+i(y+1).$

So the modulus condition becomes

\iff \frac{|z-2i|}{|z+i|}=2$$ Hence, $$|z-2i|=2|z+i|.$$ --- 3. **Convert into Cartesian form** Now, $$|z-2i|=\sqrt{x^2+(y-2)^2},$$ $$|z+i|=\sqrt{x^2+(y+1)^2}.$$ Thus, $$\sqrt{x^2+(y-2)^2}=2\sqrt{x^2+(y+1)^2}.$$ Squaring both sides, $$x^2+(y-2)^2=4\bigl(x^2+(y+1)^2\bigr).$$ Expand: $$x^2+y^2-4y+4=4x^2+4y^2+8y+4.$$ Bring all terms to one side: $$0=3x^2+3y^2+12y.$$ Divide by $3$: $$x^2+y^2+4y=0.$$ Complete the square in $y$: $$x^2+(y+2)^2=4.$$ --- 4. **Identify the circle** The equation $$x^2+(y+2)^2=2^2$$ represents a circle of radius $2$ and centre $$(0,-2).$$ --- 5. **Match with the options** So the correct option is: **A: $(0,-2)$** --- 6. **Compare with stored correct answer** Stored correct answer = **A**. This matches our derived answer.

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