Finding the modulus and argument of a complex number

[Bashyboy]

Hello,

I have the point $-1 + 2i$, for which I am asked to find the modulus and argument. The modulus was simple enough, but I am having difficulty finding the angle. The point is located in the 4th quadrant, and so I need to make certain that I calculate an angle in the range $(\frac{3 \pi}{2}, 2 \pi )$, if I wish to measure counterclockwise rotations. Here is what I did:

$\theta = \arctan(-2) = - \arctan(2)$.

So, clearly I have a negative angle, which is reasonable. However, I would like to find a positive angle, but I can't seem to wrap my head around the concept of coterminal angles and calculating angles based upon that principle.

 

[Mark44]

Hello,

I have the point $-1 + 2i$, for which I am asked to find the modulus and argument. The modulus was simple enough, but I am having difficulty finding the angle. The point is located in the 4th quadrant

No it isn't.

, and so I need to make certain that I calculate an angle in the range $(\frac{3 \pi}{2}, 2 \pi )$, if I wish to measure counterclockwise rotations. Here is what I did:

$\theta = \arctan(-2) = - \arctan(2)$.

So, clearly I have a negative angle, which is reasonable. However, I would like to find a positive angle, but I can't seem to wrap my head around the concept of coterminal angles and calculating angles based upon that principle.

Add $2\pi$ radians to get the positive angle.

 

[Bashyboy]

Whoops, you are certainly correct. I meant to type in 1-2i. The principle is the same, however.

 

[Mark44]

Arctan(-2) will give you a negative angle. Add $2\pi$ to get a positive angle. For example, if you got $-\pi/6$, adding $2\pi$ would give you $11\pi/6$ as the positive angle.

 

[Bashyboy]

I thought you just said that I should add an angle of $\pi$.

 

[Mark44]

I confused myself. You thought the angle was in the 4th quadrant (it isn't - it's in the 2nd quadrant). Adding π would get your angle to the second quadrant, which isn't what you want. I edited my earlier post.

 

[WWGD]

You can also think of the line going through the origin and through the point 1-2i (equiv., (1, -2) in Cartesian ). This may help you visualize how to choose the right angle.

 

[Ray Vickson]

Whoops, you are certainly correct. I meant to type in 1-2i. The principle is the same, however.

For $0 < \alpha < \pi/2$ the angles $-\alpha$ and $2 \pi - \alpha$ both describe the same point in the third quadrant.