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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2021 · 16 March · Morning Shift · Mathematics · Complex Number

Let z and w be two complex numbers such that w = z ¯ z - 2z + 2,...

Let zz and ww be two complex numbers such that w=zz¯−2z+2,∣z+iz−3i∣=1w = z\overset{¯}{z} - 2z + 2, \left|\dfrac{z + i}{z - 3i}\right| = 1 and Re(w)Re \left(w\right) has minimum value. Then, the minimum value of n∈Nn \in N for which wn{w}^{n} is real, is equal to _______.Let z and w be two complex numbers such that w = z ¯ z - 2z + 2, |(z + i)/(z - 3i)| = 1 and Re (w) has minimum value.

Correct answer

0

Step-by-step solution

Let z and w be two complex numbers such that w = z ¯ z - 2z + 2, |(z + i)/(z - 3i)| = 1 and Re (w) has minimum value.

ω=zz¯−2z+2\omega = z\overset{¯}{z} - 2z + 2

\left|\dfrac{z + i}{z - 3i}\right| = 1</p><p></p><p>⇒ \left|z + i\right| = \left|z - 3i\right|

⇒x2+(y+1)⇒ {x}^{2} + {\left(y + 1\right)}^{2} = {x}^{2} + {\left(y - 3\right)}^{2}

⇒y=1⇒ y = 1

⇒z=x+i,x∈R⇒ z = x + i, x \in R

ω=(x+i)(x−i)\omega = \left(x + i\right) \left(x - i\right) - 2 \left(x + i\right)+2+ 2

=x2+1−2x−2i+2= {x}^{2} + 1 - 2x - 2i + 2

Re(ω)Re \left(\omega\right) = {x}^{2} - 2x + 3 = {\left(x - 1\right)}^{2} + 2

For min \left(Re\left(\omega\right)\right), x = 1</p><p></p><p>⇒ \omega = 2 - 2i = 2 \left(1 - i\right) = 2\sqrt{2}{e}^{- i\dfrac{\pi}{4}}</p><p></p><p>{\omega}^{n} = {\left(2\sqrt{2}\right)}^{n}{e}^{- i\dfrac{n\pi}{4}}</p><p>Forrealandminimumvalueof</p><p>For real and minimum value ofn,</p><p>,</p><p>n = 4

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