1- i -1- i 2-1- i -1- i 2 is equal toA- -2B- 2C- 2 iD- -2 i

We have, ( 1+i/1 i )^2 + ( 1 i/1+i )^2 = 2i/ 2i + 2i/2i [∵ (1 + i)^2 = 1 + i^2 + 2i and (1 i^2) = 1 + i2 2i = 2i] = 1 1 = 2 ...

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Greatest Binomial Coefficients

1+ i /1- i 2+...

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${(\frac{1 + i}{1 - i})}^{2} + {(\frac{1 - i}{1 + i})}^{2}$ is equal to

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${(\frac{1 + i}{1 - i})}^{2} + {(\frac{1 - i}{1 + i})}^{2}$ is equal to

{(\frac{1 + i}{1 - i})}^{2} + {(\frac{1 - i}{1 + i})}^{2}

{(\frac{1 + i}{1 - i})}^{2} + {(\frac{1 - i}{1 + i})}^{2}

Let I(n) = 2cos nx $\forall n \in N$ , then I (1).I(n+1) - I(n) is equal to

\infty \sum i = 1 \frac{i^{2}}{2^{i}}

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