Polar Form of Complex Number/Examples/-root 6 - root 2 i

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Example of Polar Form of Complex Number

The complex number $-\sqrt 6 - \sqrt 2 i$ can be expressed as a complex number in polar form as $\polar {2 \sqrt 2, \dfrac {7 \pi} 6}$.


Proof

-root 6 - root 2 i.png
\(\ds \cmod {-\sqrt 6 - \sqrt 2 i}\) \(=\) \(\ds \sqrt {\paren {-6}^2 + \paren {-2}^2}\) Definition of Complex Modulus
\(\ds \) \(=\) \(\ds \sqrt {6 \times 2}\)
\(\ds \) \(=\) \(\ds 2 \sqrt 2\)


Then:

\(\ds \map \cos {\map \arg {-\sqrt 6 - \sqrt 2 i} }\) \(=\) \(\ds \dfrac {-\sqrt 6} {2 \sqrt 2}\) Definition of Argument of Complex Number
\(\ds \) \(=\) \(\ds -\frac {\sqrt 3} 2\)
\(\ds \leadsto \ \ \) \(\ds \map \arg {-\sqrt 6 - \sqrt 2 i}\) \(=\) \(\ds \dfrac {5 \pi} 6 \text { or } \dfrac {7 \pi} 6\) Cosine of $150 \degrees$, Cosine of $210 \degrees$


\(\ds \map \sin {\map \arg {-\sqrt 6 - \sqrt 2 i} }\) \(=\) \(\ds \dfrac {-\sqrt 2} {2 \sqrt 2}\) Definition of Argument of Complex Number
\(\ds \) \(=\) \(\ds -\frac 1 2\)
\(\ds \leadsto \ \ \) \(\ds \map \arg {-\sqrt 6 - \sqrt 2 i}\) \(=\) \(\ds \dfrac {7 \pi} 6 \text { or } \dfrac {11 \pi} 6\) Sine of $210 \degrees$, Sine of $330 \degrees$


Hence:

$\map \arg {-\sqrt 6 - \sqrt 2 i} = \dfrac {7 \pi} 6$

and hence the result.

$\blacksquare$


Sources