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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Complex Number

If z= 3 2 +(i)/(2), i= -1 , then (z^ 201 -i)^ 8 is equal to

If z=32+i2,i=−1z=\frac{\sqrt{3}}{2}+\frac{i}{2}, i=\sqrt{-1}, then (z201−i)8\left(z^{201}-i\right)^{8} is equal to

Options

  1. A1
  2. B256
  3. C-1
  4. D0

Correct answer

B. 256

Step-by-step solution

z=32+i2=eiπ/6z = \frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6}.
z201=ei⋅201π/6=ei⋅33.5π=ei(33π+π/2)z^{201} = e^{i \cdot 201\pi/6} = e^{i \cdot 33.5\pi} = e^{i(33\pi + \pi/2)}.
Since ei⋅33π=eiπ=−1e^{i \cdot 33\pi} = e^{i\pi} = -1, we get z201=−1⋅eiπ/2=−iz^{201} = -1 \cdot e^{i\pi/2} = -i.
z201−i=−i−i=−2iz^{201} - i = -i - i = -2i.
(z201−i)8=(−2i)8=28⋅i8=256⋅1=256(z^{201} - i)^8 = (-2i)^8 = 2^8 \cdot i^8 = 256 \cdot 1 = 256.
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