Question

What is the trigonometric form of `12+4i`?

Answer 1

`12+4i=4sqrt10(cosalpha+isinalpha)`, where `tanalpha=1/3`

Explanation:

`a+ib` can be written in trigonometric form `re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)`,
where `r=sqrt(a^2+b^2)`.

Hence `12+4i=sqrt(12^2+4^2)[cosalpha+isinalpha]`

= `sqrt160[cosalpha+isinalpha]`

= `4sqrt10[cosalpha+isinalpha]`,

where `cosalpha=12/(4sqrt10)=3/sqrt10`

and `sinalpha=4/(4sqrt10)=1/sqrt10`

or `tanalpha=(1/sqrt10)/(3/sqrt10)=1/3`

Answer 2

`4sqrt10(cos(0.322)+isin(0.322))`

Explanation:

Given a complex number z = x + yi , this can be written in trig. form as.

`z=x+yi=r(costheta+isintheta)`

where `r=sqrt(x^2+y^2)`

and `theta=tan^-1(y/x)`

here x = 12 and y = 4

`rArrr=sqrt(12^2+4^2)=sqrt160=4sqrt10`

and `theta=tan^-1(4/12)≈0.322" radians or"18.43^@`

`rArr12+4i=4sqrt10(cos(0.322)+isin(0.322))`or

`12+4i=4sqrt10(cos(18.43)^@+isin(18.43)^@)`