Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.1

Solution

Given z1=2+i, z2=32i, z3=1+3i.First find z1z3:z1z3=(2+i)(1+3i)=2+6i+i+3i2=2+7i3=1+7iNow divide by z2 using conjugate (3+2i):z1z3z2=1+7i32i3+2i3+2i=(1+7i)(3+2i)(32i)(3+2i)Expand numerator:(1+7i)(3+2i)=32i+21i+14i2=3+19i14=17+19iSimplify denominator:(32i)(3+2i)=32+22=9+4=13Write in a+ib form:17+19i13=1713+1913i\begin{aligned} & \boxed{\text{Given } z_1=2+i,\ z_2=3-2i,\ z_3=1+3i.} \\ \\ & \boxed{\text{First find } z_1z_3:} \\ \\ & z_1z_3=(2+i)(1+3i) \\ & = 2+6i+i+3i^2 \\ & = 2+7i-3 = -1+7i \\ \\ & \boxed{\text{Now divide by } z_2 \text{ using conjugate } (3+2i):} \\ \\ & \frac{z_1z_3}{z_2}=\frac{-1+7i}{3-2i}\cdot\frac{3+2i}{3+2i} \\ & = \frac{(-1+7i)(3+2i)}{(3-2i)(3+2i)} \\ \\ & \boxed{\text{Expand numerator:}} \\ \\ & (-1+7i)(3+2i) = -3-2i+21i+14i^2 \\ & = -3+19i-14 = -17+19i \\ \\ & \boxed{\text{Simplify denominator:}} \\ \\ & (3-2i)(3+2i)=3^2+2^2=9+4=13 \\ \\ & \boxed{\text{Write in } a+ib \text{ form:}} \\ \\ & \frac{-17+19i}{13} = -\frac{17}{13}+\frac{19}{13}i \end{aligned}