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( (√(2))/(2) \right)\\ \qquad 3√(2)\\ b=6sin(315^o)\\ \qquad 6\left( -(√(2))/(2) \right)\\ \qquad -3√(2) \end{cases} \\\\\\ z=(~3√(2)~,~-3√(2)~)\implies z=3√(2)-3√(2)~i`

Answer 2

The answer is:

`√(6)(\cos 315\°+i\sin 315\°)=√(3)-i√(3)`

Explanation:

The complex number is given as:

`√(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\°=(1)/(√(2))`

( since, we have:

`\cos (360\°-\theta)=\cos \theta` )

`\sin 315\°=\sin (360\°-45\°)\\\\i.e.\\\\\sin 315\°=-\sin 45\°=-(1)/(√(2))`

( since, we have:

`\sin (360\°-\theta)=-\sin \theta` )

Now, the expression is simplified as follows:

`√(6)(\cos 315\°+i\sin 315\°)=√(6)((1)/(√(2))-i(1)/(√(2)))`

i.e.

`√(6)(\cos 315\°+i\sin 315\°)=√(6)* (1)/(√(2))-i√(6)* (1)/(√(2))`

i.e.

`√(6)(\cos 315\°+i\sin 315\°)=√(3)-i√(3)`