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UPSC NDA I 2016 Mathematics Solved Paper

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Question Numbers: 52-53

For the next two (2) items that follow:

Let z be a complex number satisfying

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Question : 52 of 120

Marks: +1, -0

| \frac{z-4}{z-8} | =1

| \frac{z}{z-2} | =\frac{3}{2}

What is | z | equal to?

6
12
18
36

Solution:

Concept:

For any complex number

z = x + iy

, the modulus is

|z| = \sqrt{x^2 + y^2}

.

If

z_1

and

z_2

are complex numbers with

z_2 \neq 0

, then

\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}

.

Explanation:

Given:

\left|\frac{z-4}{z-8}\right| = 1

and

\left|\frac{z}{z-2}\right| = \frac{3}{2}

.

Let

z = x + iy

.

From the first equation:

\left|\frac{z-4}{z-8}\right| = 1 \Rightarrow \frac{|z-4|}{|z-8|} = 1

.

Thus

|z-4| = |z-8|

.

Substitute:

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

.

Square both sides:

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

.

Simplify:

x^2 - 8x + 16 = x^2 - 16x + 64

.

\Rightarrow -8x + 16 = -16x + 64 \Rightarrow 8x = 48

? Let's recalc: Actually

x^2 - 8x + 16 = x^2 - 16x + 64

gives

-8x + 16 = -16x + 64 \Rightarrow 8x = 48 \Rightarrow x = 6

. Correct.

Now use the second equation:

\left|\frac{z}{z-2}\right| = \frac{3}{2} \Rightarrow \frac{|z|}{|z-2|} = \frac{3}{2}

.

So

2|z| = 3|z-2|

.

Substitute

z = 6 + iy

:

2\sqrt{6^2 + y^2} = 3\sqrt{(6-2)^2 + y^2} = 3\sqrt{4^2 + y^2}

.

Square both sides:

4(36 + y^2) = 9(16 + y^2)

.

\Rightarrow 144 + 4y^2 = 144 + 9y^2 \Rightarrow 4y^2 = 9y^2 \Rightarrow 5y^2 = 0 \Rightarrow y = 0

.

Thus

z = 6 + 0i = 6

.

Hence

|z| = \sqrt{6^2 + 0^2} = 6

.

Answer:

Option A: 6.

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