JEE Main 2026 22 January Morning Shift
Exam JEE Main
Year 2026
Date 22 January
Shift Morning Shift
Subject Mathematics
Chapter Complex Number JEE Main 2026 · 22 January · Morning Shift · Mathematics · Complex Number
Let α= -1+i 3 2 and β= -1-i 3 2 , i= -1 . If (7-7 α+9 β)^ 20 +(9+7...
Let
α = − 1 + i 3 2 \alpha=\frac{-1+i \sqrt{3}}{2} α = 2 − 1 + i 3 and
β = − 1 − i 3 2 , i = − 1 \beta=\frac{-1-i \sqrt{3}}{2}, i=\sqrt{-1} β = 2 − 1 − i 3 , i = − 1 . If
( 7 − 7 α + 9 β ) 20 + ( 9 + 7 α − 7 β ) 20 + ( − 7 + 9 α + 7 β ) 20 + ( 14 + 7 α + 7 β ) 20 = m 10 (7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20}+(14+7 \alpha+7 \beta)^{20}=m^{10} ( 7 − 7 α + 9 β ) 20 + ( 9 + 7 α − 7 β ) 20 + ( − 7 + 9 α + 7 β ) 20 + ( 14 + 7 α + 7 β ) 20 = m 10 , then
m m m is
_ _ _ _ _ \_\_\_\_\_ _____
Step-by-step solution
α = ω \alpha = \omega α = ω ,
β = ω 2 \beta = \omega^2 β = ω 2 (cube roots of unity).
Let
z 1 = 7 − 7 ω + 9 ω 2 z_1 = 7-7\omega+9\omega^2 z 1 = 7 − 7 ω + 9 ω 2 ,
z 2 = 9 + 7 ω − 7 ω 2 z_2 = 9+7\omega-7\omega^2 z 2 = 9 + 7 ω − 7 ω 2 ,
z 3 = − 7 + 9 ω + 7 ω 2 z_3 = -7+9\omega+7\omega^2 z 3 = − 7 + 9 ω + 7 ω 2 .
Calculating:
z 1 = 6 − 8 i 3 z_1 = 6-8i\sqrt{3} z 1 = 6 − 8 i 3 ,
z 2 = 9 + 7 i 3 z_2 = 9+7i\sqrt{3} z 2 = 9 + 7 i 3 ,
z 3 = − 15 + i 3 z_3 = -15+i\sqrt{3} z 3 = − 15 + i 3 .
All have
∣ z k ∣ = 228 |z_k| = \sqrt{228} ∣ z k ∣ = 228 .
Verify:
z 2 = z 1 ω z_2 = z_1\omega z 2 = z 1 ω ,
z 3 = z 1 ω 2 z_3 = z_1\omega^2 z 3 = z 1 ω 2 .
z 1 20 + z 2 20 + z 3 20 = z 1 20 ( 1 + ω 20 + ω 40 ) = z 1 20 ( 1 + ω 2 + ω ) = 0 z_1^{20}+z_2^{20}+z_3^{20} = z_1^{20}(1+\omega^{20}+\omega^{40}) = z_1^{20}(1+\omega^2+\omega) = 0 z 1 20 + z 2 20 + z 3 20 = z 1 20 ( 1 + ω 20 + ω 40 ) = z 1 20 ( 1 + ω 2 + ω ) = 0 .
Fourth term:
14 + 7 ω + 7 ω 2 = 14 + 7 ( − 1 ) = 7 14+7\omega+7\omega^2 = 14+7(-1) = 7 14 + 7 ω + 7 ω 2 = 14 + 7 ( − 1 ) = 7 . So
7 20 = 4 9 10 7^{20} = 49^{10} 7 20 = 4 9 10 .
Total
= 0 + 4 9 10 = m 10 ⇒ m = 49 = 0 + 49^{10} = m^{10} \Rightarrow m = 49 = 0 + 4 9 10 = m 10 ⇒ m = 49 .
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