1. When multiplying and dividing two complex numbers ${z}_{1}$ and  ${z}_{2}$,  the moduli and arguments of the numbers are connected by the following sets of relationships:

$|{z}_{1}{z}_{2}|=|{z}_{1}||{z}_{2}|andarg({z}_{1}{z}_{2})=arg({z}_{1})+arg({z}_{2})$

$|\frac{{z}_{1}}{{z}_{2}}|=\frac{|{z}_{1}|}{|{z}_{2}|}andarg(\frac{{z}_{1}}{{z}_{2}})=arg({z}_{1})-arg({z}_{2})$

When the numbers are given in modulus-argument form as ${z}_{1}={r}_{1}(cos{ \theta }_{1}+isin{ \theta }_{1})$ and  ${z}_{2}={r}_{2}(cos{ \theta }_{2}+isin{ \theta }_{2})$, where ${r}_{1}=|{z}_{1}|,{r}_{2}=|{z}_{2}|,{ \theta }_{1}=arg{z}_{1}and{ \theta }_{2}=arg{z}_{2}$,  these relationships mean that

${z}_{1}{z}_{2}={r}_{1}{r}_{2}(cos({ \theta }_{1}+{ \theta }_{2})+isin({ \theta }_{1}+{ \theta }_{2}))$

$\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}(cos({ \theta }_{1}-{ \theta }_{2})+isin({ \theta }_{1}-{ \theta }_{2}))$

Use those results to work out the following, giving your answers in modulus-argument form:

(i) $5(cos\frac{ \pi }{3}+isin\frac{ \pi }{3})\times4(cos\frac{ \pi }{2}+isin\frac{ \pi }{2})$

(ii) $12(cos\frac{ \pi }{2}+isin\frac{ \pi }{2})\div3(cos\frac{3 \pi }{4}+isin\frac{3 \pi }{4})$

(iii) $\sqrt{3}(cos\frac{ \pi }{6}+isin\frac{ \pi }{6})\times\sqrt{6}(cos(-\frac{3 \pi }{4})+isin(-\frac{3 \pi }{4}))$

(iv) $\frac{6.25(cos3.05+isin3.05)}{0.5(cos1.42+isin1.42)}$