How do you multiply (73i)(5i) in trigonometric form?

1 Answer
Mar 8, 2017

(73i)(5i)=2377(cosθ+isinθ), where θ=tan1(1116)

Explanation:

Let us write the two complex numbers in polar coordinates and let them be

z1=r1(cosα+isinα) and z2=r2(cosβ+isinβ)

Here, if two complex numbers are a1+ib1 and a2+ib2 r1=a21+b21, r2=a22+b22 and α=tan1(b1a1), β=tan1(b2a2)

Their multipication leads us to

{r1r2}{(cosα+isinα)(cosβ+isinβ)} or

{r1r2}{(cosαcosβ+i2sinαsinβ)+i(cosαsinβ+cosβsinα)) or

{r1r2}{(cosαcosβsinαsinβ)+i(cosαsinβ+cosβsinα)) or

(r1r2)(cos(α+β)+isin(α+β)) or

z1z2 is given by (r1r2,(α+β))

So for multiplication of complex number z1 and z2 , take new angle as (α+β) and modulus os r1r2 of the modulus of two numbers.

Here 73i can be written as r1(cosα+isinα) where r1=72+(3)2=58 and α=tan1(37)

and 5i can be written as r2(cosβ+isinβ) where r2=52+(1)2=26 and β=tan1(15)

and z1z2=58(26)(cosθ+isinθ), where θ=α+β

Hence, tanθ=tan(α+β)=tanα+tanβ1tanαtanβ=37+(15)1(37×(15))=22353235=2232=1116.

Hence, (73i)(5i)=58×26(cosθ+isinθ)

= 2377(cosθ+isinθ), where θ=tan1(1116)