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Question: Additive inverse of \((1 - i)^{n} = 2^{n},\) is....

Additive inverse of (1−i)n=2n,(1 - i)^{n} = 2^{n}, is.

A

n=n =

B

−1- 1

C

(1+i)5×(1−i)5(1 + i)^{5} \times (1 - i)^{5}

D

None of these

Answer

(1+i)5×(1−i)5(1 + i)^{5} \times (1 - i)^{5}

Explanation

Solution

If −1+1−1+1+.....- 1 + 1 - 1 + 1 + ..... is the additive inverse of (2n+1)(2n + 1) then 1+i2+i3−i6+i8=1−1−i+1+1=2−i1 + i^{2} + i^{3} - i^{6} + i^{8} = 1 - 1 - i + 1 + 1 = 2 - i ⇒ ∑n=1200in=i+i2+i3+....+i200=i(1−i200)1−i\sum_{n = 1}^{200}{i^{n} = i + i^{2} + i^{3} + .... + i^{200} = \frac{i(1 - i^{200})}{1 - i}}, =i(1−1)1−i=0= \frac{i(1 - 1)}{1 - i} = 0

⇒ ∑n=113(in+in+1)\sum_{n = 1}^{13}{(i^{n} + i^{n + 1})}, =(i+i2+i3+....+i13)+(i2+i3+....+i14)= (i + i^{2} + i^{3} + .... + i^{13}) + (i^{2} + i^{3} + .... + i^{14})

=i(1−i13)1−i+i2(1−i13)1−i=i(1−i1−i)+i2(1−i)(1−i)= \frac{i(1 - i^{13})}{1 - i} + \frac{i^{2}(1 - i^{13})}{1 - i} = i\left( \frac{1 - i}{1 - i} \right) + \frac{i^{2}(1 - i)}{(1 - i)}The additive inverse of =i+i2=i−1= i + i^{2} = i - 1is (i−1i+1×i−1i−1)n=(−2i−2)n=in\left( \frac{i - 1}{i + 1} \times \frac{i - 1}{i - 1} \right)^{n} = \left( \frac{- 2i}{- 2} \right)^{n} = i^{n}

Trick : Since z=i[1+3+5+....+(2n+1)]z = i^{\lbrack 1 + 3 + 5 + .... + (2n + 1)\rbrack}.