# Surds in polar form of imaginary number

1. Feb 22, 2013

### Aerospace93

1. The problem statement, all variables and given/known data
Find the polar form for zw by first putting z and w into polar form.
z=2√3-2i, w= -1+i

2. Relevant equations
Tan-1(-√3/3)= 5∏/6

3. 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 22, 2013

### CAF123

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

3. Feb 22, 2013

### HallsofIvy

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

4. Feb 22, 2013

### 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?

5. Feb 22, 2013

### CAF123

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

Do you see how the other argument arises?

6. Feb 22, 2013

### Aerospace93

hun? im dont 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, im not sure how to work myself around this yet

7. Feb 22, 2013

### HallsofIvy

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

8. Feb 22, 2013

### 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?

9. Feb 22, 2013

### 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)] ?

10. Feb 22, 2013

### Aerospace93

could someone confirm this for me? thanks

11. Feb 22, 2013

### 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.