Express $\frac 1{1+\frac 1{1-\frac 1{1+i}}}$ in the form $a+bi$, where $a$ and $b$ are real numbers.
In case you have trouble reading that, here's a simpler version:
1/(1+1/(1-1/(1+i))).
Simplify the following:
1/(1/(-1/(i + 1) + 1) + 1)
Multiply numerator and denominator of (-1)/(i + 1) by 1 - i:
1/(1/((-(-i + 1))/((i + 1) (-i + 1)) + 1) + 1)
(1 + i) (1 - i) = 1×1 + 1 (-i) + i×1 + i (-i) = 1 - i + i + 1 = 2:
1/(1/(1 - ((-i + 1))/2) + 1)
Put each term in 1 - (-i + 1)/2 over the common denominator 2: 1 - (-i + 1)/2 = 2/2 + (-1 + i)/2:
1/(1/(2/2 + (i - 1)/2) + 1)
2/2 + (i - 1)/2 = ((1 i - 1) + 2)/2:
1/(1/((2 - 1 + i)/2) + 1)
Multiply the numerator of 1/((2 - 1 + i)/2) by the reciprocal of the denominator. 1/((2 - 1 + i)/2) = (1×2)/(2 - 1 + i):
1/(2/(2 - 1 + i) + 1)
2 - 1 + i = (2 - 1) + i = 1 + i:
1/(2/(i + 1) + 1)
Multiply numerator and denominator of 2/(i + 1) by 1 - i:
1/((2 (-i + 1))/((i + 1) (-i + 1)) + 1)
(1 + i) (1 - i) = 1×1 + 1 (-i) + i×1 + i (-i) = 1 - i + i + 1 = 2:
1/((2 (-i + 1))/2 + 1)
(2 (-i + 1))/2 = 2/2×(-i + 1) = -i + 1:
1/(-i + 1 + 1)
1 + 1 - i = (1 + 1) - i = 2 - i:
1/(-i + 2)
Multiply numerator and denominator of 1/(-i + 2) by 2 + i:
(i + 2)/((-i + 2) (i + 2))
(2 - i) (2 + i) = 2×2 + 2 i - i×2 - i×i = 4 + 2 i - 2 i + 1 = 5:
(i + 2)/5 = 2/5 + i/5