Write in polar form $$4-9i$$ and $$37+18i$$.
See development and solution
Development:
To write a binomial number in polar form, we must calculate its module and its argument. By means of the formula proposed we have:
$$|4-9i|=\sqrt{4^2+9^2}=\sqrt{16+81}=\sqrt{97}$$
$$\alpha=\arctan(\dfrac{-9}{4})$$
This way: $$$ 4-9i= \sqrt{97}_{\arctan(-\frac{9}{4})}$$$
and with the second one,
$$|37+18i|=\sqrt{37^2+18^2}=\sqrt{1369+324}=\sqrt{1693}$$
$$\alpha=\arctan(\dfrac{18}{37})$$
This way: $$$ 37+18i= \sqrt{1693}_{\arctan(-\frac{18}{37})}$$$
Solution:
$$ 4-9i= \sqrt{97}_{\arctan(-\frac{9}{4})} \ $$ and $$ \ 37+18i= \sqrt{1693}_{\arctan(-\frac{18}{37})}$$