if (^3+i)100=2^99(a+ib),then show that a^2+b^2=4
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Answer:
Since ω=
3
−1+i
3
=
2
i
2
+i
3
⇒
3
+i=
i
2ω
=−2ωi
Thus (
3
+i)
100
=(−2ωi)
100
=2
100
⋅ω
100
i
100
=2
100
ω(1)=2
100
(
2
−1+i
3
)=2
99
(−1+i
3
)
So a=−1 and b=
3
Hence a
2
+b
2
=1+3=4
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