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# Can you get "imaginary" areas? watch

1. I was just doing some complex number work and I was thinking if you could get complex areas.

For example, . If you integrated "y" between -10 and 0 you could get a numerical answer in terms of "i". And then this value changes, if you change the function so if you had or .

Do these imaginary values for the area have any meaning? Or is it just a mistake from not defining the function properly or something?
2. I'd be interested to know the answer to this too I can't think of any obvious reason why you shouldn't be able to do calculus with imaginary numbers. But I know absolutely nothing about this so I'm looking forwards to the good people of TSR enlightening me
3. This area of maths is to do with complex analysis. I haven't done any really so I don't want to answer your specific question, but yeah you can integrate imaginary numbers and stuff. Here's a good intro:

http://people.math.gatech.edu/~cain/winter99/ch4.pdf

The Riemann integral stuff they're going on about is just the normal integral you're used to - the idea behind that integral is you can say that value of the integral can be overestimated using rectangles and it can also be underestimated using rectangles too. So that L <= I <= U where L is the lower estimate, I is the integral and U is the upper estimate. If you use more and more rectangles to approximate the integral, then you should usually find that L and U become closer and closer. In the limit as you get an infinite number of rectangles, you should have L = U, so that L <= I <= U becomes L = I = U. That's the idea behind an integral and the article appears to be talking about extending it with complex numbers.
4. You need to take care about what actually means for x between -10 and 0. Does (-1)^(1/2) = i or i? Of course this issue arises when we look at the square root of positive values, i.e. 3 and -3 are both viable candidates for the square root of 9. But by convention, to ensure that f(x) = x^(1/2) defines a function on 0 <= x < infinity, we choose the positive value - the value > 0. But there is no such useful* ordering system for complex numbers. So we say x^(1/2) = exp(log(x^(1/2)) = exp(1/2*log(x)), and then do what is choosing a branch¬ of the complex logarithm.^ Now x^(1/2) (which cosmetically should be z^(1/2) since we're venturing in the realm of complex values) makes sense, and we may proceed. As z^(1/2) has an anti-derivative, explicitly 2/3*z^(3/2) where z^(3/2) is similarly defined as exp(3/2*log(z)) with a branch of the logarithm chosen, finding the value of the integral is quite simple and no more difficult than a real value case.

*there is lexicographic order, but it's not useful in complex analysis.

^ this problem arises since the function exp(z) is infinitely-many-to-one. Take 1 for example, exp(0) = 1, but exp(2pi*i) = exp(4pi*i) = exp(2k*pi*i) = 1 for any integer k. So we need to 'cut' our preimage to define an inverse function.

¬ the commonest choice of branch is the principal branch, given by log(z) = log(|z|) + i*arg z where arg z lives in (-pi, pi]. It's chosen because it agrees with the standard real value of log for the positive numbers.

This is a lot of rambling about being careful with complex values and logarithms and suchlike, and I've not really answered your question. But I hope it's useful nonetheless.
5. I seem to recall something to do with phasors (something to do with expressing sine waves in terms of complex numbers by considering oscillations, IIRC) where integrating a phasor (which is a complex number) gives another phasor. Haven't really studied phasors at all so I'll go no further but I suppose that sounds like an application.
6. Thanks for the replies, very interesting
7. I'm guessing you can - not on a normal graph though, because if you think about it, the normal 2-d x,y graphs are where x and y are both real, so the imaginary space is somewhere else not on the graph, which I think means its more complicated than just integrating.

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