schoolAccedmy
Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.1

(ii) (−2+3i)21+i\frac{(-2 + 3i)^2}{1 + i}

Solution

First expand (−2+3i)2:(−2+3i)2=(−2)2+2(−2)(3i)+(3i)2=4−12i+9i2=4−12i−9=−5−12iNow divide by (1+i) using conjugate (1−i):−5−12i1+i⋅1−i1−i=(−5−12i)(1−i)(1+i)(1−i)Expand numerator:(−5−12i)(1−i)=−5+5i−12i+12i2=−5−7i−12=−17−7iSimplify denominator:(1+i)(1−i)=12+12=2Write in a+ib form:−17−7i2=−172−72i\begin{aligned} & \boxed{\text{First expand } (-2+3i)^2:} \\ \\ & (-2+3i)^2 = (-2)^2 + 2(-2)(3i) + (3i)^2 \\ & = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i \\ \\ & \boxed{\text{Now divide by } (1+i) \text{ using conjugate } (1-i):} \\ \\ & \frac{-5-12i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(-5-12i)(1-i)}{(1+i)(1-i)} \\ \\ & \boxed{\text{Expand numerator:}} \\ \\ & (-5-12i)(1-i) = -5 + 5i - 12i + 12i^2 \\ & = -5 - 7i - 12 = -17 - 7i \\ \\ & \boxed{\text{Simplify denominator:}} \\ \\ & (1+i)(1-i)=1^2+1^2=2 \\ \\ & \boxed{\text{Write in } a+ib \text{ form:}} \\ \\ & \frac{-17-7i}{2} = -\frac{17}{2} - \frac{7}{2}i \end{aligned}
arrow_back
Previous Part
(i)
Next Part
(iii)
arrow_forward