the value of (1+i)^1000 is
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0
Answer:
lydllylydlydkydykh
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Answered by
0
Answer:
-2^500
Step-by-step explanation:
using demoviers theorem you can defined that
by (√2) ^1000(1/√2+i/√2) ^1000
- cos π =-1
- sinπ =0
- so 2 ^500(cos1000π/4 + isin 1000π/4)
- 2^500(-1)
- -2^500
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