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Transforming the complex number $z=-\sqrt{3}+3i$ into polar form will bring me to the problem to solve this two equations to find the angle $\phi$: $\cos{\phi}=\frac{\Re z}{|z|}$ and $\sin{\phi}=\frac{\Im z}{|z|}$.

For $z$ the solutions are $\cos{\phi}=-0,5$ and $\sin{\phi}=-0,5*\sqrt{3}$. Using Wolfram Alpha or my calculator I can get $\phi=\frac{2\pi}{3}$ as solution. But using a calculator is forbidden in my examination.

Do you know any (cool) ways to get the angle without any other help?

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Draw a picture and measure the angle! –  draks ... Jul 16 '12 at 9:50
    
There are very few instances where the answer will be "nice." One needs to learn them. And you probably did at some stage. Maybe the reason you did not recognize it is writing $0.5$ instead of $\frac{1}{2}$. If you had been looking for $\cos \phi=-\frac{1}{2}$, $\sin\phi=\frac{\sqrt{3}}{2}$ you might have remembered. –  André Nicolas Jul 16 '12 at 14:19

3 Answers 3

memorize sin/cos for angles $0,{\pi \over 6},{\pi \over 4},{\pi \over 3},{\pi \over 2}$ and in your examination look at the unit circle to figure out what is going on

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memorize $\sin \phi$ and calculate $\cos \phi = \sin (\phi +\pi/2)$ –  draks ... Jul 16 '12 at 10:00

You could start with known angles (s.a. multiples of 45 or 30 degrees) and work your way from there using the formulas for trigonometric half-angles, and sums of angles. If you don't remember them, use: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

For instance, in degrees, if you want cos(41), you can use the sequence: 45+120=165 165/2=82.5 82.5/2=41.25

and use the trigonometric identities to fall back from cos(45) and cos(120) to the approximation cos(41.25)

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The direct calculation is $$\arg(-\sqrt 3+ 3i)=\arctan\frac{3}{-\sqrt{3}}=\arctan (-\sqrt 3)=\arctan \frac{\sqrt 3/2}{-1/2}$$

As the other two answers remark, you must learn by heart the values of at least the sine and cosine at the main angle values between zero and $\,\pi/2\,$ and then, understanding the trigonometric circle, deduce the functions' values anywhere on that circle.

The solution you said you got is incorrectly deduced as you wrote $$\,cos\phi=-0,5\,,\,\sin\phi=-0,5\cdot \sqrt 3\,$$ which would give you both values of $\,\sin\,,\,\cos\,$ negative, thus putting you in the third quadrant of the trigonometric circle, $\,\{(x,y)\;:\;x,y<0\}\,$, which is wrong as the value indeed is $\,2\pi/3\,$ (sine is positive!), but who knows how did you get to it.

In the argument's calculation above please do note the minus sign is at $\,1/2\,$ in the denominator, since that's what corresponds to the $\,cos\,$ in the polar form, but the sine is positive thus putting us on the second quadrant $\,\{(x,y,)\;:\;x<0<y\}\,$ .

So knowing that $$\sin x = \sin(\pi - x)\,,\,\cos x=-\cos(\pi -x)\,,\,0\leq x\leq \pi/2$$ and knowing the basic values for the basic angles, gives you now $$-\frac{1}{2}=-\cos\frac{\pi}{3}\stackrel{\text{go to 2nd quad.}}=\cos\left(\pi-\frac{\pi}{3}\right)=\cos\frac{2\pi}{3}$$ $$\frac{\sqrt 3}{2}=\sin\frac{\pi}{3}\stackrel{\text{go to 2nd quad.}}=\sin\left(\pi-\frac{\pi}{3}\right)=\sin\frac{2\pi}{3}$$

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