Answer:
[tex]z_1=1-(2-\sqrt{3})i=2\sqrt{2-\sqrt{3}}[cos(\frac{23\pi}{12} )+isin(\frac{23\pi}{12} )][/tex]
Step-by-step explanation:
Recall that [tex]a+bi=r(cos\theta+isin\theta)[/tex] where [tex]r=\sqrt{a^2+b^2}[/tex] and [tex]\theta=tan^{-1}(\frac{b}{a})[/tex].
Since [tex]a=1[/tex] and [tex]b=-2+\sqrt{3}[/tex], then [tex]r=\sqrt{(1)^2+(-2+\sqrt{3})^2}=2\sqrt{2-\sqrt{3}}[/tex] and [tex]\theta=tan^{-1}(\frac{-2+\sqrt{3}}{1})=-\frac{\pi}{12}=\frac{23\pi}{12}[/tex].
Therefore, [tex]z_1=1-(2-\sqrt{3})i=2\sqrt{2-\sqrt{3}}[cos(\frac{23\pi}{12} )+isin(\frac{23\pi}{12} )][/tex]
