Question

Question #f593a

Answer 1

`1/2cis((3pi)/8)`

Explanation:

Consider the 2 complex numbers.

`z_1=r_1(costheta_1+isintheta_1)`

and `z_2=r_2(costheta_2+isintheta_2)`

dividing them gives.

`(z_1)/(z_2)=(r_1(costheta_1+isintheta_1))/(r_2(costheta_2+isintheta_2))`

multiply numerator and denominator by `(costheta_2-isintheta_2)`

`(z_1)/(z_2)=(r_1(costheta_1+isintheta_1)(costheta_2-isintheta_2))/(r_2(costheta_2+isintheta_2)(costheta_2-isintheta_2))`

distribute the brackets.

`=(r_1(costheta_1costheta_2+sintheta_1sintheta_2+i(sintheta_1costheta_2-costheta_1sintheta_2)))/(r_2(cos^2theta_2+sin^2theta_2))`

`=[(r_1)/(r_2)](cos(theta_1-theta_2)+isin(theta_1-theta_2)........ (A)`

(A) is obtained using the `color(blue)"trigonometric identities"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(cos(A-B)=cosAcosB+sinAsinB)color(white)(a/a)|)))" and "`

`color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A-B)=sinAcosB-cosAsinB)color(white)(a/a)|)))`

and of course, the denominator `cos^2theta_2+sin^2theta_2=1`

The result being `color(red)(|bar(ul(color(white)(a/a)color(black)(|z_1/(z_2)|=(r_1)/(r_2)" and " arg((z_1)/(z_2))=theta_1-theta_2)color(white)(a/a)|)))`
`color(blue)"----------------------------------------------------------------"`

Here `r_1=5,theta_1=(5pi)/8" and " r_2=10,theta_2=pi/4`

`rArr(5cis((5pi)/8))/(10cis(pi/4))=1/2cis((5pi)/8-pi/4)=1/2cis((3pi)/8)`