I think it is easier to multiple (5-7i) (a+bi) and set the real part of the result equal to 148 and the imaginary part equal to zero. Then you will have a set of 2 linear equations with 2 unknowns, a and b. Simply solve those. Perhaps this takes a little bit longer, but not much.

the (14i)x (-7i) or b components didnt seem right because it feels like it wont eliminate the pesky central i components that normally cancel out with conjugation, but then when you realize that you are doubling the “a” component also– they do cancel out. It intuitively appears that the rule (a+bi)(a-bi)=(a^2 +b^2) is the only way to eliminate the central ( garbage)component but as long as the form ( xa+xbi)( a-bi) or x( a+b)( a-b) holds you will simply end up with x(a^2+b^2) Another words the distributive law even holds with conjugation and i numbers. Even fantasy fiction i must adhere to some normal conventions!

## Comments (3)

I got the conjugation, but the doubling was troubling.

I think it is easier to multiple (5-7i) (a+bi) and set the real part of the result equal to 148 and the imaginary part equal to zero. Then you will have a set of 2 linear equations with 2 unknowns, a and b. Simply solve those. Perhaps this takes a little bit longer, but not much.

the (14i)x (-7i) or b components didnt seem right because it feels like it wont eliminate the pesky central i components that normally cancel out with conjugation, but then when you realize that you are doubling the “a” component also– they do cancel out. It intuitively appears that the rule (a+bi)(a-bi)=(a^2 +b^2) is the only way to eliminate the central ( garbage)component but as long as the form ( xa+xbi)( a-bi) or x( a+b)( a-b) holds you will simply end up with x(a^2+b^2) Another words the distributive law even holds with conjugation and i numbers. Even fantasy fiction i must adhere to some normal conventions!