Question 69072: Solve (1 + i) / (1 - i)
Note how parentheses have been added to make the numerator (1+i) and the denominator (1-i) which is what I think you intended.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Any time that you have a complex number in the denominator look to change it to a real number by multiplying it by its complex conjugate ... which is to say multiply it by itself except with a change in the sign between the real and imaginary portions of the complex number.
Easier to understand if we work this problem. In the denominator there is the term 1-i. Multiply it by 1+i. (This is the complex conjugate of 1-i ... same terms only change of the sign between the terms.)
Recall from algebra that: (a-b)(a+b) = a^2 - b^2. Thats exactly what we have except a = 1 and b=i. So the product of (1-i)(1+i) = 1^2 -(i^2) = 1 - i^2.
But also recall that by definition i^2 = -1. So our multiplication which resulted in 1 - i^2 becomes 1 - (-1) when we substitute -1 for i^2. And 1 - (-1) becomes 1 +1 = 2. Therefore, after multiplying the original denominator of 1-i by its conjugate of 1+i the denominator is converted to 2.
However, we can't multiply the denominator by 1+i without also multiplying the numerator by 1+i. [In effect that is equivalent to multiplying the original fraction by (1+i)/(1+i) which is to say, multiplying the original fraction by 1.]
Since the original fraction had 1+i and the numerator, we multiply that by 1+i (which is to say that we square the numerator in this case). The result is:
(1+i)*(1+i) = 1^2 + 1i + 1i + i^2. This simplifies to 1+2i+i^2, but recall that i^2 equals -1 by definition. Therefore we substitute -1 for i^2 to get 1+2i-1. The real parts cancel and we are left with just 2i in the numerator. We now divide this by the denominator of 2 to get 2i/2 = i.
The answer is just i.
Hope this helps you to think about complex number division
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