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Question

Question: If \(2i\) then \(- 2i\)...

If 2i2i then 2i- 2i

A

2- 2

B

22

C

i592+i590+i588+i586+i584i582+i580+i578+i576+i5741=\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} - 1 =

D

(2, 1)

Answer

2- 2

Explanation

Solution

(1i)x+(1+i)y=13i( 1 - i ) x + ( 1 + i ) y = 1 - 3 i in+in+1+in+2+in+3=in[1+i+i2+i3]i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = i^{n}\lbrack 1 + i + i^{2} + i^{3}\rbrack =in[1+i1i]=0= i^{n}\lbrack 1 + i - 1 - i\rbrack = 0

Equating real and imaginary parts, we get (1+i)2=1+i2+2i=2i\because(1 + i)^{2} = 1 + i^{2} + 2i = 2iand (1i)2=1+i22i=2i(1 - i)^{2} = 1 + i^{2} - 2i = - 2i; (1+i)8+(1i)8=(2i)4+(2i)4\Rightarrow (1 + i)^{8} + (1 - i)^{8} = (2i)^{4} + ( - 2i)^{4} =24(i4+i4)= 2^{4}(i^{4} + i^{4}). Thus point is =25=32.= 2^{5} = 32..