JEE Main 2026 23 January Evening 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 = 3 2 + i 2 , i = − 1 z=\frac{\sqrt{3}}{2}+\frac{i}{2}, i=\sqrt{-1} z = 2 3 + 2 i , i = − 1 , then
( z 201 − i ) 8 \left(z^{201}-i\right)^{8} ( z 201 − i ) 8 is equal to
Step-by-step solution
z = 3 2 + i 2 = e i π / 6 z = \frac{\sqrt{3}}{2} + \frac{i}{2} = e^{i\pi/6} z = 2 3 + 2 i = e iπ /6 .
z 201 = e i ⋅ 201 π / 6 = e i ⋅ 33.5 π = e i ( 33 π + π / 2 ) z^{201} = e^{i \cdot 201\pi/6} = e^{i \cdot 33.5\pi} = e^{i(33\pi + \pi/2)} z 201 = e i ⋅ 201 π /6 = e i ⋅ 33.5 π = e i ( 33 π + π /2 ) .
Since
e i ⋅ 33 π = e i π = − 1 e^{i \cdot 33\pi} = e^{i\pi} = -1 e i ⋅ 33 π = e iπ = − 1 , we get
z 201 = − 1 ⋅ e i π / 2 = − i z^{201} = -1 \cdot e^{i\pi/2} = -i z 201 = − 1 ⋅ e iπ /2 = − i .
z 201 − i = − i − i = − 2 i z^{201} - i = -i - i = -2i z 201 − i = − i − i = − 2 i .
( z 201 − i ) 8 = ( − 2 i ) 8 = 2 8 ⋅ i 8 = 256 ⋅ 1 = 256 (z^{201} - i)^8 = (-2i)^8 = 2^8 \cdot i^8 = 256 \cdot 1 = 256 ( z 201 − i ) 8 = ( − 2 i ) 8 = 2 8 ⋅ i 8 = 256 ⋅ 1 = 256 .
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