Question

Perform the following multiplication and write the product in trigonometric form. Write the magnitude in exact form. Write the argument in radiansand round it to two decimal places if necessary.(-4 + 3D)( V2 + 2i)

Answer 1

First, calculate the result of the multiplication, as shown below

`\begin{gathered} w=(-4+3i)(√(2)+2i)=-4√(2)+i3√(2)-8i+6(-1) \\ =-4√(2)-6+i(3√(2)-8) \end{gathered}`

In general, given a complex number z=a+ib, its magnitude is given by the formula below

`\begin{gathered} z=a+ib \\ \Rightarrow||z||=√(a^2+b^2) \end{gathered}`

Then, in our case,

`\begin{gathered} \Rightarrow||w||=(-4√(2)-6)^2+(3√(2)-8)^2 \\ \Rightarrow||w||=4(17+12√(2))+2(41-24√(2)) \\ \Rightarrow||w||=5√(6) \end{gathered}`

Furthermore,

`\begin{gathered} w=r(cos\theta+isin\theta) \\ and \\ r=5√(6) \end{gathered}`

Then, finding theta,

`\begin{gathered} \Rightarrow cos\theta=(-4√(2)-6)/(5√(6)) \\ \Rightarrow\theta\approx-2.8289... \\ \Rightarrow\theta\approx-2.83 \end{gathered}`

Hence, the answer is

`w=5√(6)(cos(-2.83)+isin(-2.83))`