Answer:
\(\sqrt{3} -i\sqrt{3}\)
Step-by-step explanation:
The given complex number is,
\(\sqrt{6}(\cos 315^{\circ} + i \sin 315^{\circ})\)
we know that,
\(\sin 315^{\circ}=-\dfrac{1}{\sqrt2}\) and
\(\cos 315^{\circ}=\dfrac{1}{\sqrt2}\)
Putting the values,
\(=\sqrt{6}\left(\dfrac{1}{\sqrt2} -i\dfrac{1}{\sqrt2}\right)\)
\(=\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2} -i\dfrac{1}{\sqrt2}\right)\)
\(=\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2}\right) -i\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2}\right)\)
\(=\sqrt{3} -i\sqrt{3}\)
