Math, asked by dsn8267, 6 months ago

Q-8-Express
i/1-i
in the form of a +ib​

Answers

Answered by Raghav1112
2

Answer:

1/1-i Rationalise it

1/1-i×1+i/1+i

1+i/1-i^2

1+i/1+1

1+i/2 i^2 =-1,so-i^2 =1

1/2 + i/2

real part=1/2

imaginary part=i/2

Answered by anshikaankur311
0

Answer:

Answer:1-i/-2

Answer:1-i/-2Step-by-step explanation:

Answer:1-i/-2Step-by-step explanation:i/1-i

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it,

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-i

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2i

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2i

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2iRationalise it again because we can't leave iota in denominator

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2iRationalise it again because we can't leave iota in denominator i+1/-2i × - 2i/-2i

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2iRationalise it again because we can't leave iota in denominator i+1/-2i × - 2i/-2i-2i^2-2i/4i^2

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2iRationalise it again because we can't leave iota in denominator i+1/-2i × - 2i/-2i-2i^2-2i/4i^22-2i/-4

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2iRationalise it again because we can't leave iota in denominator i+1/-2i × - 2i/-2i-2i^2-2i/4i^22-2i/-41-i/-2

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2iRationalise it again because we can't leave iota in denominator i+1/-2i × - 2i/-2i-2i^2-2i/4i^22-2i/-41-i/-2-1/2 is real part

Answer:1-i/-2Step-by-step explanation:i/1-iRationalise it, i/1-i× 1-i/1-ii-i^2/1+i^2-2ii+1/1-1-2i ( as i^2=-1)i+1/-2iRationalise it again because we can't leave iota in denominator i+1/-2i × - 2i/-2i-2i^2-2i/4i^22-2i/-41-i/-2-1/2 is real part i/2 is an imaginary part

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