Question

Let z and ω be two complex numbers such that ω = `zbarz - 2z + 2,|(z + i)/(z - 3i)|` = 1 and Re(ω) has minimum value. Then, the minimum value of n∈N for which ωⁿ is real, is equal to ______.

पर्याय

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Answer 1

Let z and ω be two complex numbers such that ω = `zbarz - 2z + 2,|(z + i)/(z - 3i)|` = 1 and Re(ω) has minimum value. Then, the minimum value of n∈N for which ωⁿ is real, is equal to 4.

Explanation:

Given that ω = `zbarz - 2z + 2,|(z + i)/(z - 3i)|` = 1  ...(i)

Using `|z_1/z_2|` = 1 ⇒ |z₁| = |z₂|,

We can write `|(z + i)/(z - 3i)|` = 1 as,

|z + i| = |z – 3i|  ...(ii)

Put z = x + iy in equation (ii)

|x + iy + i| = |x + iy – 3i|

⇒ |x + (y + 1)i| = |x + (y – 3)i|  ...(iii)

Since, |z| = `sqrt(x^2 + y^2)`,

So from equation (iii)

`sqrt(x^2 + (y + 1)^2) = sqrt(x^2 + (y - 3)^2)`

⇒ x² + y² + 2y + 1 = x² + y² – 6y + 9

⇒ 8y = 8

⇒ y = 1  ...(iv)

Now using equation (i)

ω = `(x + iy)(bar(x + iy)) - 2(x + iy) + 2`

Using `z.barz = |z|^2`, we can write

ω = x² + y² – 2(x + iy) + 2

⇒ ω = (x² + y² – 2x + 2) + 2yi

ω = (x² – 2x + 3) – 2i; y = 1  ...(v)

Now, Re(ω) = x² – 2x + 3

⇒ Re(ω) = 2 + (x – 1)²

Re(ω) will be the minimum when x = 1.

Hence z = 1 + i; x = y = 1

Now using equation (v)

ω = 1 – 2 + 3 – 2i

⇒ ω = 2 – 2i 

⇒ ω = 2(1 – i) = ω

⇒ `2sqrt(2)(1/sqrt(2) - 1/sqrt(2)i)`

ω = `2sqrt(2)(1/sqrt(2) + (-1/sqrt(2))i)`

⇒ ω = `2sqrt(2)(cos  π/4 + sin  (-π/4)i)`

ω = `2sqrt(2)(cos(-π/4) + sin  (π/4)i)`

⇒ ω =  `2sqrt(2)e^(i(π/4)`

⇒ ω^π = `(2sqrt(2))^n e^(i((nπ)/4)`

ωⁿ is real and minimum when n = 4.