Asked by henro
Express square root of(1-i/1+i) in form of a+bit where a and b are real numbers
Answers
Answered by
Steve
recall that (a+bi)(a-bi) = a^2+b^2
(1-i)/(1+i) * (1-i)/(1-i)
= (1-i)^2 / (1+1)
= (1-2i+i^2)/2
= (1-2i-1)/2
= -2i/2
= -i
or, 0 - i
(1-i)/(1+i) * (1-i)/(1-i)
= (1-i)^2 / (1+1)
= (1-2i+i^2)/2
= (1-2i-1)/2
= -2i/2
= -i
or, 0 - i
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