Which of the following is equivalent to \dfrac{26-13i}{8+i} ? (Note: i=\sqrt{-1})




Comments (3)

I understood the entire question but got (208 – 26i – 104i + 13i^2) over the correct denominator. Why is negative 13i times negative i equal to negative 13i^2?

Hey, Mike. This is a nice problem. It seems to me that you can save a fair amount of work by factoring out the 13 in the numerator: 13 (2 – i) ………..
This helps/works because we know that the denominator will be 65 (because it’s always the first number squared plus the coefficient-of-i squared) and 13 is a factor of 65. That lets us reduce the fraction we’re working with to (2-i)(8-i)/5, giving us (15 – 10i)/5 = 3 – 2i
This is especially helpful since CB test writers seem to like to put these complex denominator problems in Section 3/Non-calculator and making the arithmetic easier from the jump is pretty helpful there.

Leave a Reply