1- i -1- i 2-1- i -1- i 2 is equal toA- 2 iB- -2 iC- -2D- 2

The correct option is C 2 (1+i/1 i)^2 + (1 i/1+i)^2 = 2i/ 2i + ( 2i/2i) = 2 ...

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Properties of Iota

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

Question

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

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-2i

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-2

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

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

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

if i = $\sqrt{- 1}$ , then $1 + i^{2} + i^{3} + i^{6} + i^{8}$ is equal to

I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{⊖}

\frac{- d [I_{2}]}{d t} = - \frac{1}{2} \frac{d [S_{2} O_{3}^{2 -}]}{d t}

\frac{- d [I_{2}]}{d t} = - 2 \frac{d [S_{2} O_{3}^{2 -}]}{d t}

\frac{- d [I_{2}]}{d t} = - 2 \frac{d [I^{⊖}]}{d t} \frac{d [S_{2} O_{3}^{2 -}]}{d t}

\frac{d [S_{4} O_{6}^{2 -}]}{d t} = \frac{1}{2} \frac{d [I^{⊖}]}{d t}

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