Given:
z1 = 6(cos80 + isin80)
z2 = 8(cos140 + isin140)
Let's solve for the following:
(a) z1/z2
To solve for z1/z2, we have:
[tex]\begin{gathered} \frac{z_1}{z_2}=\frac{6(\cos 80+i\sin 80)}{8(\cos 140+i\sin 140)} \\ \\ \frac{z_1}{z_2}=\frac{3(\cos 80+i\sin 80)}{4(\cos 140+i\sin 140)} \end{gathered}[/tex]Solving further:
[tex]\frac{z_1}{z_2}=\frac{0.521+2.954i}{-3.064+2.571i}[/tex]Multiply the denominator and numerator by the conjugate:
[tex]\begin{gathered} \frac{0.521+2.954i}{-3.064+2.571i}\times\frac{-3.064-2.571i}{-3.064-2.571i} \\ \\ \frac{(0.521+2.954i)(-3.064-2.571i)}{(-3.064+2.571i)(-3.064-2.571i)} \\ \\ =\frac{6-10.392i}{16} \\ \\ =\frac{1}{16}\ast\frac{6-10.392i}{1} \end{gathered}[/tex]Let's write in trigonometric form:
[tex]\begin{gathered} 0.0625(6-10.392i) \\ \\ 0.0625(6)+0.0625(-10.392i) \\ \\ 0.375-0.649i \\ \\ \lvert z\rvert=\sqrt[]{(-0.649)^2+(}0.375)^2 \\ \\ \lvert z\rvert=\sqrt[]{0.562}=0.749 \\ \\ \tan ^{-1}(\frac{-0.649}{0.375})=-60^o \end{gathered}[/tex]Therefore, the answer in trigonometric form is:
[tex]0.749(\cos (-60^0)+i\sin (-60))[/tex]Part b.
[tex]z_1z_2=(6(\text{cos}80+i\sin 80))\times(8(\cos 140+i\sin 140))[/tex]Thus, we have:
[tex]\begin{gathered} z_1z_2=\mleft(1.042+5.909i\mright)\times(-6.128+5.142i) \\ \\ \end{gathered}[/tex]Apply the FOIL method:
[tex]1.042(-6.128+5.142i)+5.909i(-6.128+5.142i)[/tex]Apply distributive property:
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