a)

The polar coordinate $\left[\begin{array}{cc}r& \theta \end{array}\right]$ corresponding to Cartesian coordinate $\left[\begin{array}{cc}x& \text{}y\end{array}\right]is\text{}given\text{}by[r=\sqrt{{x}^{2}\text{}+\text{}{y}^{2}}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)].$

From the given data, $[(x,\text{}y)=(-2,\text{}2)].$

Substitute $[(x,\text{}y)=(-2,\text{}2)]\in [r=\sqrt{{x}^{2}\text{}+\text{}{y}^{2}}]$ and obtain the value of r.

$r=\sqrt{{2}^{2}\text{}+\text{}{(-2)}^{2}}$

$=\pm \sqrt{8}$

$=\pm \text{}2\sqrt{2}$

Similary, calculate the value of $\left[\theta \right]$ as follows.

$\theta ={\mathrm{tan}}^{-1}(-\frac{2}{2})$

$={\mathrm{tan}}^{-1}(-1)$

$=\frac{3\pi}{4}\text{}+\pi n,\text{}n\text{}\in \mathbb{Z}$

For $[r=2\sqrt{2}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}n=0]the\text{}polar\text{}c\text{}in\text{}coordinate\text{}is\left[\begin{array}{cc}2\sqrt{2}& \frac{3\pi}{4}\end{array}\right].$

For $[r=2\sqrt{2}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}n=2\mathrm{\setminus}]the\text{}polar\text{}c\text{}in\text{}coordinate\text{}is\left[(2\sqrt{2},\frac{11\pi}{4})\mathrm{\setminus}\right].$

For $[r=2\sqrt{2}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}n=\text{}-2\mathrm{\setminus}]the\text{}polar\text{}c\text{}in\text{}coordinate\text{}is\left[(2\sqrt{2},\text{}-\frac{5\pi}{4})\mathrm{\setminus}\right]$.

For $[r=2\sqrt{2}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}n=\text{}-1\mathrm{\setminus}]the\text{}polar\text{}c\text{}in\text{}coordinate\text{}is\left[(-2\sqrt{2},\text{}-\frac{\pi}{4})\mathrm{\setminus}\right]$

Hence the given stament is true.

b) Note that the Cartesian coordinates corresponding to polar coordinates

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