My college lecturer gave me this as a "bonus" question. Apparently we're supposed to be able to solve it with just a-level maths knowledge? i) Sketch the graph of x^2 + y^2 = 0 ii) Sketch the graph of [(x^2 - 1)(x^2 - 4)]^2 + [(y^2 - 1)(y^2 - 4)]^2 = 0 All the graphs i've encountered so far have been of the form y = something in x. How do you sketch a graph where x and y are mixed up?

The first one is either wrong or your teacher wants you to suffer. x^2 + y^2 = r^2 where r is a constant is the graph of a circle with a radius equal to r. r being equal to 0 means your graph is in the complex plane and can't actually be drawn on normal cartesian axes. The second one is very tricky. What you want to do is to find solutions that make your function 0. There are various solutions which you put on the graph. You are gonna get is a very weird formation which is basically two squares and a point in the middle. You have to use derivatives to find turnings points, maximums and minimums. When you find them, try to understand how the function behaves. There is little you can do further. You can use Wolfram for a general idea: www.wolframalpha.com/ even though I encourage you to try solving some stuff by yourself. As you can see, both of your functions do not have a graph which is indicative of their very weird behaviour. You can also try Microsoft Mathematics for more detailed equations and maybe finding a solution to these. I personally believe it's a mistake in writing the questions, there is no way these functions can be drawn normally and especially the first one which is not even on the cartesian axes.

OH, i understand the first part. Cause if either x or y weren't zero then the total would be positive and not equal to zero. So the graph is just a point? And the second is 16 points? I'm still not too clear on how you got the random points for part ii though..

If the left part (the x-part) is zero, then the right part (the y-part) also has to be zero for the entire equation to be zero. Now, lets say x = -1. Then the entire left part is zero. There are four possibilities for the right part to be zero; y = -2 or -1, and y = 2 or = 1 (because y^2 is the same for both sets). Now, this is the case for every one of the 4 possibilities of x for which the x-part is zero. Thus, 4*4 means there are 16 different possibilities.

When r=0 it means geometrically that it's just a point. (I supposed that's what his teacher wanted) Anyway, if he ever wanted to graph it, it's just a lineal complex function. Or just a point in cartesian plane. x²+y²=0 -> y=ix -> x=iy So basically there wouldn't be any problem if he knew that sqrt(-1) = i In the case of the second one, it's not that hard. We already know all the 16 roots so the graph (in cartesian plane) is just 16 points.

It seems peculiar to me that the teacher would just ask for single points unless he wanted to trick the students into overthinking it. After all, the problem does not facilitate any graph drawing methods to be found and especially the second in which there is no function continuity to be found outside of complex numbers. Drawing on the complex plane would be fine, but at OP's level I'd assume it's not something they have covered and would trouble most students.

Yes, our teacher is quite the fan of making his students suffer with weird questions LOL. On our first day in college he gave us an infinite nested radical to solve. We were like "what in sweet hell is this"