Polar form help

[duhherro]

How would you convert (1,-root 3) into polar form? Just need  help with this :S, can't get the books answer [2,pi/3]

[brightsky]

picture it on an argand diagram. (1, -sqrt(3)) would be in the 4th quadrant. the distance of the point from the origin would be sqrt(1^2 + 3) = 2. the angle it makes with the positive direction of the x-axis would be tan^(-1)(-sqrt(3)) = -pi/3.

[duhherro]

picture it on an argand diagram. (1, -sqrt(3)) would be in the 4th quadrant. the distance of the point from the origin would be sqrt(1^2 + 3) = 2. the angle it makes with the positive direction of the x-axis would be tan^(-1)(-sqrt(3)) = -pi/3.

Haha thanks thought but could you provide me with just the working out? we learn argand diagrams later :S

[Phy124]

For now:

The short version: (these are very simple once you know the logistics)

Do as follows:

Let x = x coordinate on argand diagram, y = y coordinate on argand diagram





^both simple trigonometry

*Note using short method... if it is in quadrant 2 then you have to add and if it is in quadrant 3 then you have to minus

The polar form you're looking for is , so

For when you get into complex numbers: (I'm not sure whether you've learnt these, as I didn't do GMA or Spesh )

on an argand diagram is in polar form.

The long version: (only really used for differing understanding, you may be made to do this if you take up Engineering early into complex numbers -.-)





(1)

(2)

Divide 1 by 2;





Which gets us to what we had above:



Sine is negative, whilst cosine is positive, so it is in the fourth quadrant, therefore we can leave it as this.

To find r, combine equations 1 and 2 and square both sides



Simplify that, take r^2 out as the common factor, etc. and you end up with:



This derives what we used earlier:



As above, the polar form you're looking for is , so

edit: Put answer in the form needed

[brightsky]

picture it on an argand diagram. (1, -sqrt(3)) would be in the 4th quadrant. the distance of the point from the origin would be sqrt(1^2 + 3) = 2. the angle it makes with the positive direction of the x-axis would be tan^(-1)(-sqrt(3)) = -pi/3.

Haha thanks thought but could you provide me with just the working out? we learn argand diagrams later :S

argand diagrams are basically just cartesian planes, except instead of x and y axis, you have Re(z) and Im(z) axis. when you express a complex number in cartesian form z = x + yi, you are essentially saying that the x-coordinate, or real coordinate, of z is x, and the y-coordinate, or imaginary coordinate, of z is y. so we can plot the point on a cartesian plane/argand diagram as (x,y). how do you convert from cartesian to polar form? well expressing a complex number in polar form means putting it in the form z = r cis (theta), where r is the DISTANCE THE POINT IS FROM THE ORIGIN, and theta is THE ANGLE IT MAKES WITH THE POSTIVE DIRECTION OF THE X-AXIS (though there is a condition: -pi < theta =< pi).

if you're given the point (1, -sqrt(3)), how do you find the distance of that point from the origin? distance formula/pythagoras! how do you find the angle it makes with the positive direction of x-axis? simple trig!

[Aurelian]

Guys, I think you're overlooking the fact that polar form doesn't necessarily have to relate to complex numbers - as might be the case here  (unless I'm mistaken!). Perhaps we shouldn't be confusing him with complex numbers (are they even in the Gen A course? I'm not sure as I didn't do it haha).

The OP has given a point consisting entirely of real numbers. That point is (1, -sqrt(3)). The OP is actually just looking for an answer in the simple form of (magnitude, angle).

OP; just think about a normal Cartesian plane, with an x-axis and a y-axis. Ignore this stuff with "cis" and the like. Look at the above point - where is it located? Hopefully you can see that it is located in the 4th quadrant (the bottom right quadrant).

Imagine drawing a line from the origin to that point. We could specify the above point by giving a length of this line, and what angle that line makes with the x-axis.

So what's the length of this line? Well, really, we can quite easily see a right triangle, so we can apply pythagoras; a^2 + b^2 = c^2. Thus, in this case:





So we have the length - but what angle does this line segment make with the x-axis? Well, we can find the smaller angle quite easily, just by considering a normal right triangle with lengths sqrt(3) (y-axis) and 1 (x-axis). Hopefully when you draw this out it becomes clear that

tan(theta) = sqrt(3)/1 = sqrt(3)
Therefore, theta = pi/3.

However, we've gone *clockwise* around the plane. Positive angles are taken going anticlockwise, so really our angle is -pi/3, since we've gone pi/3 radians around the plane in the negative direction.

Thus, the length of the line is 2, and the angle that line makes with the x-axis is -pi/3. So, we could identify this point as being (2, -pi/3), or perhaps (-pi/3, 2).

Which is the form given in the opening post

[pi]

+1, I think he means polar graph form (hence the co-ordinates in the answer are enclosed in square brackets - as is the norm in polar graphs) rather than complex polar form