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.

## 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?

Thanks for this comment—you caught a typo in the solution! See attached image for the correct steps.

https://uploads.disquscdn.com/images/90e2813637cb8bb6349add2b6c9272f5d41d8bc3ffbeabd34d33d9c741b3d455.jpg

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.