Question

Write the complex number in the form a + bi.

square root of six(cos 315° + i sin 315°)

Answer 1

$a=\sqrt6\cdot \cos 315=\sqrt 6 \cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt{12}}{2}=\dfrac{2\sqrt3}{2}=\sqrt3\\ b=\sqrt6\cdot \sin 315=\sqrt 6 \cdot\left(-\dfrac{\sqrt2}{2}\right)=-\dfrac{\sqrt{12}}{2}=-\dfrac{2\sqrt3}{2}=-\sqrt3\\\\ a+bi=\sqrt3-\sqrt3i$

Answer 2

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,

$=\sqrt{6}\left(\dfrac{1}{\sqrt2} -i\dfrac{1}{\sqrt2}\right)$

$=\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2} -i\dfrac{1}{\sqrt2}\right)$

$=\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2}\right) -i\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2}\right)$

$=\sqrt{3} -i\sqrt{3}$