Question

How do you express the complex number in trigonometric form: `-7-5i`?

Answer 1

`-7-5i = sqrt74(cos2.52-isin2.52)`

Explanation:

Any `a+bi` can be rewritten in mod-arg form as `r(cosvartheta +isinvartheta)` where `r = sqrt(a^2+b^2)` and `vartheta = arctan (b/a)`

`|-7-5i| = sqrt(5^2+7^2)=sqrt74`

`"arg"(-7-5i) = arctan ((-5)/(-7)) = -2.52`

`therefore -5-7i -= sqrt74 (cos(-2.52)+isin (-2.52)) = sqrt74(cos2.52-isin2.52)`

Answer 2

`sqrt74(cos(2.52)-isin(2.52))`

Explanation:

`"to convert from "color(blue)"complex to trig. form"`

`"that is "x+yitor(costheta+isintheta)" using"`

`•color(white)(x)r=sqrt(x^2+y^2)`

`•color(white)(x)theta=tan^-1(y/x)color(white)(x)-pi < theta<=pi`

`"here "x=-7" and "y=-5`

`rArrr=sqrt((-7)^2+(-5)^2)=sqrt74`

`-7-5i" is in the third quadrant so we must ensure "theta`
`"is in the third quadrant"`

`theta=tan^-1(5/7)=0.62larrcolor(blue)" related acute angle"`

`rArrtheta=-pi+0.62=-2.52larrcolor(blue)" in third quadrant"`

`rArr-7-5itosqrt74(cos(-2.52)+isin(-2.52))`

`=sqrt74(cos(2.52)-isin(2.52))`