Question

Write 1/i - 1/(1-i) in the form a+bi, where a, b is an element of R

Answer 1

Try putting the terms over a common denonimator:
\[(1-i-i)\div(i(1-i))\]

Answer 2

I'll assume this is complex numbers and i^2 = -1

get the common denominator i(1 -i)

((1-i) -i)/i(1-i) = 1 - 2i/(i -i^2)

= (1 - 2i)/(i + 1)

rationalise the denominator

(1 -2i)/(i +1) x (i -1)/(i -1) = (i -2i^2 + 1 -2i)/(i^2 - 1)

= (3 -i)/(-2)

= -3/2 + i/2

a = -3/2 and b = 1/2

Answer 3

Grouping like terms and multipying out:
\[(1-2i)\div(i+1)\]

Answer 4

Multiplying numerator and denominator by (i-1) gives:
1/2(3i+1)=1.5i+0.5=0.5+1.5i

Answer 5

Thanks guys, this question confuses me. I understand that both needs a common denominator. In my textbook I have a different example:
\[(7-2i)/(2-i)\] = \[((7-2i)(2+i))/((2-i)(2+i))\] and then you multiply to solve.

Can you explain again, how to get the common denominator using the complex conjugate of the denominator?

Answer 6

well basically its like adding 1/2 and 1/3 you need to multiply the denominators to add the numerators after cross multiplying...

once you get everything over a single denominator then you can simplify and rationalise

and I think my solution has arithmetic error... and kropots is correct

Answer 7

Sorry my solution has a negative sign left out. My answer should be :
-0.5-1.5i

Answer 8

Thanks guys, I agree with kropot's answer as well. Thanks campbell for explanations. Would like to give both medals but can't, sorry