1-i-1-i2-1-i-1-i2 is equal toA- 2 iB- -2 i2C- -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-i2+1-i/...

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

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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