[tex]\bf \qquad \textit{division of two complex numbers} \\\\ \cfrac{r_1[cos(\alpha)+isin(\alpha)]}{r_2[cos(\beta)+isin(\beta)]}\implies \cfrac{r_1}{r_2}[cos(\alpha - \beta)+isin(\alpha - \beta)] \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf \begin{cases} z1=9[cos(225^o)+i~sin(225^o)]\\\\ z2=3[cos(45^o)+i~sin(45^o)] \end{cases}\implies \cfrac{z1}{z2}\implies \cfrac{9[cos(225^o)+i~sin(225^o)]}{3[cos(45^o)+i~sin(45^o)]} \\\\\\ \cfrac{9}{3}[cos(225^o-45^o)+isin(225^o-45^o)]\implies 3[cos(180^o)+i~sin(180^o)] \\\\\\ 3[(-1)+i~(0)]\implies -3+0i[/tex]
Answer:
-1
Step-by-step explanation:
Use DeMoivre's Theorem for division of complex polar numbers. Set it up like this:
[tex]\frac{9(cos225+sin225i)}{3(cos45+sin45i)}[/tex]
The rule is
[tex]\frac{z_{1} }{z_{2} }cis(\theta_{1}-\theta _{2} )[/tex]
Our z1 is 9 and our z2 is 3, so the division of those gives you 3. The subtraction of the angles 225 - 45 = 180, therefore:
3 cis 180° = 3(cos 180 + sin 180 i)
The cos of 180 is -1 and the sin of 180 is 0, so 3(-1 + 0) = -3
