Express the complex number in trigonometric form.

-6i

Question

Express the complex number in trigonometric form.

-6i

sidsiddhartha · Answer

first let $$-6i=a+ib$$
so what's ur a and b?

anonymous · Answer

I don't know. I don't understand this.

sidsiddhartha · Answer

look here $$a=real part $$
and 
$$b=imaginary part$$
so a=0
and b=-6 do u get it?

anonymous · Answer

Yes. 
so -6 = 0 + -6i?

sidsiddhartha · Answer

no we are just comparing -6i with a+ib and we get b=-6 as a=0 all right 
yes or no?

anonymous · Answer

okay. yes

sidsiddhartha · Answer

now if u wanna Express any complex number in trigonometric form then u have write it like this
$$z= a + ib = -6i = R(\cosΘ + isinΘ)$$
where $$R = √(a²+b²) = √(0+-6²) = 6$$
and 
$$Θ = \arctan(b/a) = \arctan(b/0) = \arctan(\infty)=π/2 $$
so far all good?

anonymous · Answer

yes

anonymous · Answer

Does that mean it is 6( cos 90 +i sin 90)?

sidsiddhartha · Answer

yup very good

anonymous · Answer

YAY! Thank you!

sidsiddhartha · Answer

now a very essential thing is left look coordinates of (a,b) is=(0,-6) is'nt it

sidsiddhartha · Answer

so u have to and pi=180 degree more with each angle so it will become$$6[\cos(180+ 90) +i \sin (180+90)]=6(\cos270+isin270)$$ that's ur answer