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Minimum |z| When |z - 3 - 2i| = |z + 2i| | JEE Advanced Complex Numbers

JEE Maths question with a full step-by-step solution.

Question

|z - 3 - 2i| = |z + 2i|

, where

z

is a complex number, then the minimum value of

|z|

will be

\dfrac12

\dfrac45

\dfrac{7}{10}

\dfrac{9}{10}

correct

Solution

Step 1: Put

z = x + iy

and write both sides as distances.

\left|(x-3) + (y-2)i\right| = \left|x + (y+2)i\right| .

Step 2: Square both sides to clear the square roots.

(x-3)^{2} + (y-2)^{2} = x^{2} + (y+2)^{2} .

Step 3: Expand.

x^{2} - 6x + 9 + y^{2} - 4y + 4 = x^{2} + y^{2} + 4y + 4 .

Step 4: Cancel

x^{2}

and

y^{2}

and tidy up.

-6x + 9 - 4y = 4y \implies 6x + 8y = 9 ,

z

moves on a straight line. Step 5: Interpret

|z|

. It is the distance of the point

(x,y)

from the origin, so its minimum is the perpendicular distance from the origin to this line. Step 6: Apply the perpendicular-distance formula to

6x + 8y - 9 = 0

from

(0,0)

|z|_{\min} = \frac{\left|6(0) + 8(0) - 9\right|}{\sqrt{6^{2}+8^{2}}} = \frac{9}{\sqrt{100}} = \frac{9}{10} .

Answer: (4).

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