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

\(\displaystyle{z}={1.73}{e}^{{{2.53}{i}}}\)

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

\(\displaystyle{z}={1.73}{e}^{{{2.53}{i}}}\)