Question

How do you multiply `(-7+3i)(1+3i)` in trigonometric form?

Answer 1

`(-7+ 3 i)(1+ 3 i) =24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i)`

Explanation:

Let `Z=a+i b ; Z=-7+ 3 i ; a=-7 ,b = 3` ;

`Z=-7+ 3 i` is in `2` nd quadrant.

Modulus `|Z|=sqrt(a^2+b^2)=(sqrt((-7)^2+ 3^2)) =sqrt 58`

`tan alpha =|b/a|= 3/7 or alpha =tan^-1(3/7) ~~ 0.4049`

`theta` is on `2` nd quadrant `:. theta=pi-0.4049`

`:. theta~~ 2.7367`. Argument , `theta ~~2.7367:.`

In trigonometric form expressed as

`r(cos theta+isintheta) = sqrt58(cos 2.74+i sin 2.74)`

`Z=1+ 3 i` is in `1` st quadrant.

Modulus `|Z|=sqrt(a^2+b^2)=(sqrt(1^2+ 3^2)) =sqrt 10`

`tan alpha =|b/a|= 3/1 or alpha =tan^-1(3) ~~ 1.249`

`theta` is on `1` st quadrant `:. theta=1.249`

Argument , `theta ~~1.249:.`

In trigonometric form expressed as

`r(cos theta+isintheta) = sqrt 10 (cos 1.25+i sin 1.25)`

`(-7+ 3 i)(1+ 3 i) =`

`sqrt58(cos 2.74+i sin 2.74) * sqrt 10 (cos 1.25+i sin 1.25)~~`

`sqrt58 * sqrt 10 ( cos (2.74+1.25) + i sin(2.74+1.25) ~~`

`24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i)` [Ans]