Find real numbers a and b such that (1+i/i) - (3/4-i) = a + bi.?

Answer 1

what we need is to express (1+i) / i - 3/(4-i) in the standard form

multiply (1+i) / i by -i / -i to get (-i +1)
multiply 3/(4-i) by (4+i)/(4+i) to get (12+3i) / (16+1) = (12+3i)/17

so the expression is (-i +1) - (12/17 + 3i/17)
= (1-12/17) + i ( -1 - 3/17) (rearranging the terms)
= 5/17 + i ( -20/17)

so a=5/17 and b = -20/17

Answer 2

(1 + i/i) would be 1+1 = 2 . Is that a typo?

If not, a = 1 1/4 and b = 1

Answer 3

Do you mean (1+i) all over i minus 3 over (4-i)?

If so multiply top & bottom of the first fraction by(4-i) & top & bottom of the second by i to get common denom.

(4 + 3i - i^2) - 3i all over i(4-i)

which simplifies to 5 over 4i - i^2 , or 5 over 4i + 1

Finally multiply top and bottom by (4i - 1) giving a rational denominator. You can finish it from there I hope.