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(1+i)^2024+(1+i)^2024
Asked by spurthipuli | 27 Mar, 2024, 07:58: AM
Expert Answer
If we write above complex number in " r exp(iθ) " form , where r is magnitude and θ is argument ,
Then we have
since cos(2nπ) = 1 and sin(2nπ) = 0 , where n is an integer , above expression becomes
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