Answer (Detailed Solution Below)

Option 2 : i
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Detailed Solution

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Calculation:

Let z = \(\rm {1 + i\over1- i}\)

Multiply 1 + i on both numerator and denominator

z = \(\rm {1 + i\over1- i}\times{1 + i\over1+ i}\)

z = \(\rm (1+i)^2\over1^2-i^2\)

z = \(\rm 1+2i+i^2\over1+1\)

z = \(\rm 2i\over2\)= i

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