Given w=2(cos150∘+isin150∘)w=2(cos150) and z=–sqrt{2}+i Find the polar form of z

Question
Given \(\displaystyle{w}={2}{\left({\cos{{150}}}∘+{i}{\sin{{150}}}∘\right)}{w}={2}{\left({\cos{{150}}}\right)}{\quad\text{and}\quad}{z}=–\sqrt{{{2}}}+{i}\)
Find the polar form of z

Answers (1)

2020-11-09
We don't actually need ww in this case.
Remember that the polar form of a complex number is
\(\displaystyle{z}={r}{e}^{{{i}θ}}{o}{s}{l}{a}{s}{h}\)
where rr is the norm of z and θ angle it makes with the xx axis.
[graph]
Using a little trigonometry (form a right triangle using the negative xx axis, the arrow shown and a straight line segment between them), we know that
\(\displaystyle{r}=\sqrt{{√{2}^{{{2}}}+{1}^{{{2}}}}}=\sqrt{{{3}}}\)
and that
\(\displaystyle{\tan{{\left(π−{o}{s}{l}{a}{s}{h}\right)}}}={\frac{{{1}}}{{\sqrt{{{2}}}}}}\)
\(\displaystyleπ−{o}{s}{l}{a}{s}{h}={\arctan{{\frac{{{1}}}{{\sqrt{{{2}}}}}}}}\)
\(\displaystyle{o}{s}{l}{a}{s}{h}=\pi−{\arctan{{\frac{{{1}}}{{\sqrt{{{2}}}}}}}}\)
Hence the polar form of z is
\(\displaystyle{b}\otimes{e}{d}{\left\lbrace{z}=\sqrt{{{3}}}{e}^{{{\left[\pi-{\arctan{{\left({1}&#{x}{2}{F},\sqrt{{{2}}}\right)}}}\right]}{i}}}\right\rbrace}\)
which is approximately
\(\displaystyle{z}={1.73}{e}^{{{2.53}{i}}}\)
0

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