Question

How do you divide `(-4-9i)/(6-2i)` in trigonometric form?

Answer 1

`" "`
`color(blue)(z=(sqrt(970))/20)color(blue)([cos 264^@ + i* sin 264^@]`

Explanation:

`" "`
Let `color(red)(Z_1 = r_1(Cos theta_1 + i*sin theta_1)`

Let `color(red)(Z_2 = r_2(Cos theta_2 + i*sin theta_2)`

`color(blue)((Z_1/Z_2) = (r_1/r_2)[cos (theta_1 - theta_2)+i*sin(theta_1 - theta_2)]`

Consider the problem given:

`color(green)(Z_1 = (-4-9i)` and

`color(green)(Z_2 = (6-2i)`

In complex numbers,

we know that `i=sqrt(-1) and i^2=(-1)`

`Z_1/Z_2=(-4-9i)/(6-2i)`

Multiply and divide by the Conjugate of the denominator to simplify.

`Z_1/Z_2=(-4-9i)/(6-2i)*color(red)((6+2i)/(6+2i)`

`rArr [-24-8i-54i-18(i^2)]/(36-(2i)^2`

`rArr [-24-62i+18]/[(36+4)]`

`rArr (-62i-6)/40`

`rArr (2(-31i-3))/(2*20)`

`rArr (cancel 2(-31i-3))/(cancel 2*20)`

`rArr (-3-31i)/20`

`color(blue)( :. Z_1/Z_2= -(3/20)-(31/20)i`

Express `color(blue)(r` in terms of `color(blue)(a and b`.

Using Pythagoras Theorem,

`r^2=(a^2+b^2)`

`r=sqrt(a^2+b^2)`

`|z|=a+bi`

`|z| = sqrt(a^2+b^2`

`sin(theta)=b/r`

`rArr color(red)( b=r*sin(theta)`

`cos(theta) = a/r`

`rArr color(red)( a=r*sin(theta)`

We already have

`color(blue)( Z_1/Z_2= -(3/20)-(31/20)i`

`|z|=sqrt((-3/20)^2+(-31/20)^2`

Simplifying you get

`rArr sqrt(970/400`

`rArr sqrt(970)/20`

To find `color(red)(theta`

`theta = tan^-1[(-31/20)/(-3/(20)]]`

Using the calculator to simplify, you get

`theta ~~ 84.47245985^@`

`:. theta ~~ 84.5^@`

You have to add 180 to the angle when you recognize that the angle lies in the third quadrant.

Hence `theta ~~84.5^@ + 180^@`

`rArr theta ~~ 264^@`

Hence, the final representation of `Z` will be

`color(red)(Z=sqrt(970)/20[cos(264^@)+i*sin(264^@)]`

Hope it helps.