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Friday, 1 August 2014

CHAPTER 2 - COMPLEX NUMBERS - The Complex Plane

Complex numbers are points in the plane


In the same way that we think of real numbers as being points on a line, it is natural to identify a complex number z=a+ib with the point (a,b) in the cartesian plane. Expressions such as ``the complex number z'', and ``the point z'' are now interchangeable.
We consider the a real number x to be the complex number x+ 0i and in this way we can think of the real numbers as a subset of the complex numbers. The reals are just the x-axis in the complex plane.
The modulus of the complex number za + ib now can be interpreted as the distance from z to the origin in the complex plane.
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Since the hypotenuse of a right triangle is longer than the other sides, we have
displaymath16
for every complex number z.
We can also think of the point zaib as the vector (a,b). From this point of view, the addition of complex numbers is equivalent to vector addition in two dimensions and we can visualize it as laying arrows tail to end. (Picture)
We see in this way that the distance between two points z and w in the complex plane is |z-w|.

The identity is called the parallelogram law because if we think of z and w as vectors (or points) in the plane, then it tells us that the sum of the squares of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of its diagonals.
To prove the identity, just write
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displaymath23

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The last equality follows from tex2html_wrap_inline27 .
Similarly, we have
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Adding these gives the identity

We know the inequality when n=1 and when n=2 by the last exercise. We will show that the truth of the inequality for n=k implies it for n=k+1 when k is any integer. That will finish the proof. This is an example of proof by induction.
By the triangle inequality (in the simplest case n=2),
displaymath32
So the inductive hypothesis that
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implies
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which is the triangle inequality for the case nk+1.
 Describe geometrically, the set of all complex numbers z which satisfy the following condition
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Solution:
Since |z-1| >0, we know the set we want to describe does not contain the point z=1. By the triangle inequality, we have
displaymath18
for all z. So we want to exclude all points from the plane where the equality
displaymath22
holds. That is, we want to exclude any z whose distance from 1 is equal to 1 plus its distance to the origin. This just means we have to exclude the negative real axis and the origin. (Draw a picture.)
We can also see this algebraically. Writing zx+iy we have
displaymath32
and
displaymath34
Setting these equal gives
displaymath36
which reduces to
displaymath38
This can only hold if tex2html_wrap_inline40 and y=0.
So the set we want is the complex plane with the point z=1 and the segment tex2html_wrap_inline46 deleted.

Prove that the medians of a triangle with vertices tex2html_wrap_inline12 , tex2html_wrap_inline14 and tex2html_wrap_inline16 intersect at the point
displaymath18

Solution:
Using the previous exercise we can write the medians of the triangle as
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displaymath22
and
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These segments intersect if and only if there are real numbers tex2html_wrap_inline26 in the interval [0,1] such that
displaymath30
It's clear that tex2html_wrap_inline32 is the unique solution to this system so the point of intersection is
displaymath18

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