the value of (1+2i^5)+(1+3i)(2++1)^-1
Answers
Answered by
0
Answer:
Given that (1+i)(1+2i)(1+3i)....(1+ni)=α+iβ,
If we take modulus on both sides, ∣a×b∣=∣a∣×∣b∣
⇒
(1+1)(1+2
2
)....(1+n
2
)
=
(
α
2
+β
2
)
Squaring both sides, 2.5....(1+n
2
) equals (α
2
+β
2
)
Answered by
0
Answer:
The answer is 34
Step-by-step explanation:
hope it helps you
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