sun, I dont know why Im stuggling with a simple freakin problem. Its not even for me its for my friend who's in college algebra, but for some reaon I cant get the correct answer. \({9 - 11i \over 6i}\) I multiplied by the conjugate twice and got -11/6 + 3i/2 what did I do wrong?
i x i = -1 so: \({9 - 11i \over 6i}\) x \(i \over i\) = \({9i + 11 \over -6}\) = \({-9i \over 6}\) - \(11 \over 6\) = \({-3i \over 2}\) - \(11 \over 6\)
Wait I make mistake on the 2nd row from bottom. Now is correct. Mathman was correct Please Register or Log in to view the hidden image!
Review: Multiplication: (a + bi)(c + di) = (ab-bd)+(ad+bc)i The rule to normalize a ratio of complex numbers is to multiply top and bottom by the conjugate of the denominator. Then everything is mechanical but you have to carefully track the minus signs. So \(\begin{eqnarray} {{9-11i}\over{6i}} & = & {{9-11i}\over{6i}}\times{{-6i}\over{-6i}} \\ & = & {{(9 + (-11)i)(0 + (-6)i)}\over{(0+6i)(0+(-6)i)}} \\ & = & {{(9 \times 0 - (-11)\times(-6)) + (9 \times (-6) + 0 \times (-11))i}\over{(0 \times 0 - (6)\times(-6)) + (0 \times (-6) + 6 \times 0)i}} \\ & = & {{-66 + (-54)i}\over{36}} \\ & = & {{-11 + (-9)i}\over{6}} \\ & = & -\frac{11}{6}-\frac{3}{2}i \end{eqnarray}\)