First off, I need to tell you how to find the distance between two points when they are given in polar form, where polar form is given by (distance from origin, angle from positive x-axis). Here's a diagram:
In the above diagram, point A has the polar coordinates (r1, theta1) (yes, I know that those should be subscripts but I can't be bothered opening Word right now just to type in some subscripts and then copy-paste to Blogspot as you can't type them directly into Blogspot) and point B has the polar coordinates (r2, theta2).
The line from the origin to point A is of length r1. The line from the origin to point B is of length r2. Finally, the angle between the two lines is given by theta1 - theta2: the difference between the two angles. Now you have enough information to work out the length of AB using the cosine rule!
AB = sqrt( (r1)^2 + (r2)^2 + 2(r1)(r2)cos(theta1 - theta2))
(One day Blogspot will be able to type in not only subscripts, but square roots signs and the Greek alphabet as well... one day... or maybe I should just install a Greek keyboard on my computer?! Now that's an idea! Why did I not think of that before? Meh, can't be bothered looking for a good one right now... and I can't be bothered trying every one in the pre-installed list to see if any of them actually work...)
By the way, it doesn't matter whether you have theta1 - theta2 or theta2- theta1. The result is the same. This is because theta1 - theta2 is equal to -(theta 2 - theta1), and cos(x) = cos (-x).
Now, as for the graphs of polar equations... well, I've already written about them before in my very brief review of chapters 1-6, but I might as well just copy-paste the info here into a dedicated Polar Coordinates post.
If you have theta = (angle), then the graph is basically a line in the direction of the angle. (i.e. if the equation is theta = pi/4, then you have a line continuing at pi/4 radians, or 45 degrees, as measured anticlockwise from the positive x-axis). If it says anywhere that r doesn't have to be greater than theta, then the line extends in both directions.
If you have r = (constant), then the graph is a circle with a radius equal to the constant.
You can also get inequalities for these as well. Remember, if it's a greater than/ less than sign without the equals bit underneath, you need to draw a dotted line, not a solid line.
Oh yeah, and there are the spirals too, in the form r = k(theta), where k is a constant, r is the magnitude and theta is the angle. Normally you see these in those questions where it asks you to write the equation for the graph. Normally it helps to use the values of r at (pi/2) and/or pi to help you determine k, and, therefore, the equation.
As for complex numbers in polar form, and how to multiply and divide them, I've already written about that in my aforementioned post about Complex Numbers and Polar Coordinates.
That leaves graphing regions in the complex plane. This is pretty simple and is kind of related to the graphs of polar equations.
If you have |z| = k, where k is a constant, the graph is basically a circle with the centre as the origin and radius k. z is any number with a modulus equal to k.
If you have |z - w| = k as your equation, where w is a complex number (normally given in the form a + bi) and k is a constant, then the graph is a circle with centre w and radius k. Be careful: sometimes they'll give an equation in the form |z + w| = k. If you get this, make sure to rearrange the equation to |z - (-w)| = k, otherwise you'll end up with the centre in the wrong place.
Finally you have the annoying ones which are in the form |z - w| = |z - p|, where w and p are both complex numbers. To do these, mark points w and p on the Argand diagram. Then draw a dotted line between w and p. Finally, draw a continuous solid line that bisects the dotted line (bisecting line = a line that crosses another line in its centre and runs perpendicular to that line). Every point on the solid line should then be equidistant (i.e. of equal distance) from w and p.
And that's pretty much it for those two chapters. It kind of helps that I covered loads of stuff before...
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