Question

How do you simplify `6(cos((3pi)/4)+isin((3pi)/4))div2(cos(pi/4)+isin(pi/4))` and express the result in rectangular form?

Answer 1

`z=3i`

Explanation:

`"Given"`

`z_1=r_1(costheta_1+isintheta_1)" and "`

`z_2=r_2(costheta_2+isintheta_2)" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)((z_1)/(z_2)=(r_1)/(r_2)[cos(theta_1-theta_2)+isin(theta_1-theta_2)])color(white)(2/2)|)))`

`rArr(6[cos((3pi)/4)+isin((3pi)/4)])/(2[cos(pi/4)+isin(pi/4)`

`=6/2[cos((3pi)/4-pi/4)+isin((3pi)/4)-pi/4)]`

`=3[cos(pi/2)+isin(pi/2)]`

`=3(0+i)`

`=0+3i`

`rArrz=3i`