How do you divide 2i73i2 in trigonometric form?

1 Answer
Nov 3, 2016

2i73i2=5313(cosθ+isinθ), where θ=tan1(1720)

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 division leads us to

{r1r2}{cosα+isinαcosβ+isinβ} or

{r1r2}{cosα+isinαcosβ+isinβ×cosβisinβcosβisinβ}

(r1r2)(cosαcosβ+sinαsinβ)+i(sinαcosβcosαsinβ)(cos2β+sin2β) or

(r1r2)(cos(αβ)+isin(αβ)) or

z1z2 is given by (r1r2,(αβ))

So for division complex number z1 by z2 , take new angle as (αβ) and modulus the ratio r1r2 of the modulus of two numbers.

Here 2i7=7+2i can be written as r1(cosα+isinα) where r1=(7)2+22=53 and α=tan1(27)

and 3i2=2+3i can be written as r2(cosβ+isinβ) where r2=(2)2+32=13 and β=tan1(32)

and z1z2=5313(cosθ+isinθ), where θ=αβ

Hence, tanθ=tan(αβ)=tanαtanβ1+tanαtanβ=(27)(32)1+(27)×(32)=27+321+37=1714107=1714×710=1720.

Hence, 2i73i2=5313(cosθ+isinθ), where θ=tan1(1720)