# Complex numbers/find a and b

• February 21st 2009, 01:51 AM
$(a+bi)(1+bi) =a-b^2+(ab+b)i= 17-19i$ So $a-b^2=17$ and $ab+b=-19$. You can solve for $a$ and $b$. I've got only one real solution: $a=18$, $b=-1$.
$\left( {a + bi} \right)\left( {1 + bi} \right) = \left( {a - b^2 } \right) + \left( {ab + b} \right)i$