Question

Let \(z\) be a complex number such that \(\left|\frac{z-2 i}{z+i}\right|=2 z \neq-i\) Then \(z\) lies on the circle of radius 2 and centre

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

Answer 1

The Correct Option is D

Given the condition:

\( \frac{|z - 2i|}{|z + i|} = 2 \)

Let \(z = x + yi\), where \(x\) and \(y\) are real numbers. Substituting into the equation:

\( \frac{\sqrt{x^2 + (y - 2)^2}}{\sqrt{x^2 + (y + 1)^2}} = 2 \)

Squaring both sides to eliminate the square roots:

\( \frac{x^2 + (y - 2)^2}{x^2 + (y + 1)^2} = 4 \)

Cross-multiplying:

\( x^2 + (y - 2)^2 = 4(x^2 + (y + 1)^2) \)

Expanding both sides:

\( x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4 \)

Bringing all terms to one side:

\( x^2 + y^2 - 4y + 4 - 4x^2 - 4y^2 - 8y - 4 = 0 \)

\( -3x^2 - 3y^2 - 12y = 0 \)

\( 3x^2 + 3y^2 + 12y = 0 \)

\( x^2 + y^2 + 4y = 0 \)

To express this in standard circle form, complete the square for the \(y\)-terms:

\( x^2 + y^2 + 4y = 0 \)

\( x^2 + (y^2 + 4y + 4) = 4 \)

\( x^2 + (y + 2)^2 = 4 \)

This represents a circle with center at \((0, -2)\) and radius 2.

Thus, the correct answer is option (4)

Answer 2

The correct answer is (D) : $(0,-2)$
$(z-2i)(\overline{z}+2i)=4(z+i)(\overline{z}-i)$
$z\overline{z}+4+2i(z-\overline{z})=4(z\overline{z}+1+i(\overline{z}-z))$
$3z\overline{z}-6i(z-\overline{z})=0$
$x^2+y^2-2i(2iy)=0$
$x^2+y^2+4y=0$