Question

Write the complex number in the form a + bi. square root of six (cos 315° + i sin 315°) (2 points)

Answer 1

$\bf z=\stackrel{r}{6}[cos(\stackrel{\theta }{315^o})+i~sin(\stackrel{\theta }{315^o})]\qquad \begin{cases} a=x=rcos(\theta )\\ b=y=rsin(\theta )\\ ----------\\ a=6cos(315^o)\\ \qquad 6\left( \frac{\sqrt{2}}{2} \right)\\ \qquad 3\sqrt{2}\\ b=6sin(315^o)\\ \qquad 6\left( -\frac{\sqrt{2}}{2} \right)\\ \qquad -3\sqrt{2} \end{cases} \\\\\\ z=(~3\sqrt{2}~,~-3\sqrt{2}~)\implies z=3\sqrt{2}-3\sqrt{2}~i$

Answer 2

Answer:

The answer is:

         $\sqrt{6}(\cos 315\°+i\sin 315\°)=\sqrt{3}-i\sqrt{3}$

Step-by-step explanation:

The complex number is given as:

$\sqrt{6}(\cos 315\°+i\sin 315\°)$

Now, we know that:

$315\°=360\°-45\°$

Hence, we have:

$\cos 315\°=\cos (360\°-45\°)\\\\i.e.\\\\\cos 315\°=\cos 45\°=\dfrac{1}{\sqrt{2}}$

( since, we have:

$\cos (360\°-\theta)=\cos \theta$ )

$\sin 315\°=\sin (360\°-45\°)\\\\i.e.\\\\\sin 315\°=-\sin 45\°=-\dfrac{1}{\sqrt{2}}$

( since, we have:

$\sin (360\°-\theta)=-\sin \theta$ )

Now, the expression is simplified as follows:

$\sqrt{6}(\cos 315\°+i\sin 315\°)=\sqrt{6}(\dfrac{1}{\sqrt{2}}-i\dfrac{1}{\sqrt{2}})$

i.e.

$\sqrt{6}(\cos 315\°+i\sin 315\°)=\sqrt{6}\times \dfrac{1}{\sqrt{2}}-i\sqrt{6}\times \dfrac{1}{\sqrt{2}}$

i.e.

$\sqrt{6}(\cos 315\°+i\sin 315\°)=\sqrt{3}-i\sqrt{3}$