Surds in polar form of imaginary number

[Aerospace93]

Homework Statement

Find the polar form for zw by first putting z and w into polar form.
z=2√3-2i, w= -1+i

Homework Equations

Tan^-1(-√3/3)= 5∏/6

The Attempt at a Solution

r= √[(2√3)²+(-2)²]=4
tanθ= -2/(2√3)=-1/√3=-√3/3=> acording to above... tan^-1(-√3/3)= 5∏/6
so, in polar form z should be 4[cos(5∏/6)+isin(5∏/6)]...
However, in the markscheme as well as in another source i have (thus me deducing there was no coincidence) they've put 4[cos(-∏/6)+isin(-∏/6)].
I understand how tan(5∏/6) and tan(-∏/6) both equal(-√3/3)... though that's not the case for when it comes to cos(5∏/6) and cos(-∏/6).
So i would appreciate it if someone explained to me where i have gone wrong... and i think the question is pretty clear but if you think i haven't explained myself correctly go forth and tell me!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[CAF123]

Sketch ##z = 2 \sqrt{3} -2i## in the Argand plane. What quadrant does it lie?

 

[HallsofIvy]

Tangent is not a "one-to-one" function and so arctangent is not "well-defined". In fact, you calculator should show you that [itex]tan(5\pi/6)= tan(-\pi/6)[/itex]. Which quadrant is [itex](2\sqrt{3}, -2)[/itex] in? Where are [itex]5\pi/6[/itex] and [itex]-\pi/6[/itex]?

 

[Aerospace93]

I'm new to imaginary numbers so bare w me here...
z=2√3−2i in the Argand plane lies in the 4th quadrant...?

5π/6 lies in the 2nd Q. and −π/6 doesn't lie on any Q? π/6 lies in the first quadrant. What's the next step?

 

[CAF123]

So if your complex number lies in the 4th quadrant, what is the appropriate argument?

Do you see how the other argument arises?

 

[Aerospace93]

hun? I am don't understand what you're saying... If i look at the 4th quadrant of the trigonometric circle then wouldn't it be 11pi/6? which i guess is the reflection of the pi/6 on the x-axis?? as I've told you, I am not sure how to work myself around this yet

 

[HallsofIvy]

The trig functions, and so the polar representation of complex numbers, is "modulo [itex]2\pi[/itex]". That is [itex]-\pi/6[/itex] is exactly the same as [itex]2\pi- \pi/6= 11\pi/6[/itex].

 

[Aerospace93]

that makes sense. so i would guess that adding 11pi/6 or -pi/6 to something (pheta1 + pheta2) if i were to be multiplying to complex numbers in polar form then i would get the same result?

 

[Aerospace93]

4[cos(-∏/6)+isin(-∏/6)]*2[cos(3∏/4)+isin(3∏/4)] = 4[cos(11∏/6)+isin(11∏/6)]*2[cos(3∏/4)+isin(3∏/4)] ?

 

[Aerospace93]

could someone confirm this for me? thanks

 

[CAF123]

Yes, the two angles represent the same point, but generally we choose the argument such that ##arg(z) \in [0,\pi],##or##[0,-\pi]##

This is why the book gives the argument -pi/6 in the solution.

The reason you got 5pi/6, which also satisfies arctan(##-1/\sqrt{3})## is because there we deal with ##z = -2\sqrt{3} + 2i##. So blindly computing arctan(..) without thinking where the complex number lies causes problems.