Complex Algebra 1, Exercises - Mathematics, Exercises of Algebra

complex numbers and functions,Fermat point, contour integral, complex conjugate, arithmetic of complex numbers.

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Math 213a: Complex analysis
Problem Set #0 (15 September 2003)
A few straightforward exercises in thinking about complex numbers and functions:
1. (i) Let zbe the complex number (1 + 7i)/2. Verify that z4= (1 37i)/2, and
thus find another complex number wsuch that w4+z4= 1.
(ii) Find two complex numbers wand zsuch that wn+zn+ 1 = 0 for every positive
integer nnot divisible by 3.
2. (i) Give necessary and sufficient conditions on complex numbers z1, z2, z3to be
vertices of an equilateral triangle traversed in the “positive” (counterclockwise)
direction. [Hint: the algebra in this exercise will be much easier and more pleas-
ant if you use the R-basis {1, ω}for C, rather than the usual {1, i}, where
ω:= (1 + 3i)/2 is a cube root of unity.]
(ii) Let A, B, C be any distinct points in the Euclidean plane, and A0, B0, C 0the
points such that triangles A0BC,AB 0C,ABC0are equilateral in the same orien-
tation. Prove that with what one exception? the line segments AA0,BB0,
CC 0have the same length, and make 60angles with each other (extended if
necessary).
[It is also known that these three lines are concurrent; when the equilateral triangles are
external to 4ABC , the point of intersection is known as the Fermat point of 4ABC yes,
the same Fermat that Problem 1 should bring to mind. I do not ask that you prove the
concurrence, which cannot be easily obtained using the arithmetic of complex numbers.]
3. Let a6= 1 be a positive real number. Show that the image of the unit circle (the
set of complex numbers z=x+iy such that x2+y2= 1) under the function
z7→ z+a/z is an ellipse. What happens when a= 1, or ais a complex number?
4. Show that the area Aenclosed by a simple closed curve Cin the complex plane is
given by the contour integral
A=1
2iIC
¯z dz.
(An integral Rf dz over some path in the complex plane is interpreted as the
line integral Rf dx +iRf dy where xand yare the real and imaginary parts of
z=x+iy imagine that dz =dx +i dy. The integrand ¯zis the complex
conjugate xiy of z=x+iy. We’ll soon have a lot more to say about such
integrals.)
This problem set is due Monday, September 22, at the beginning of class.

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Math 213a: Complex analysis Problem Set #0 (15 September 2003)

A few straightforward exercises in thinking about complex numbers and functions:

  1. (i) Let z be the complex number (1 +

7 i)/2. Verify that z^4 = (1 − 3

7 i)/2, and thus find another complex number w such that w^4 + z^4 = 1. (ii) Find two complex numbers w and z such that wn^ + zn^ + 1 = 0 for every positive integer n not divisible by 3.

  1. (i) Give necessary and sufficient conditions on complex numbers z 1 , z 2 , z 3 to be vertices of an equilateral triangle traversed in the “positive” (counterclockwise) direction. [Hint: the algebra in this exercise will be much easier and more pleas- ant if you use the R-basis { 1 , ω} for C, rather than the usual { 1 , i}, where ω := (−1 +

3 i)/2 is a cube root of unity.] (ii) Let A, B, C be any distinct points in the Euclidean plane, and A′, B′, C′^ the points such that triangles A′BC, AB′C, ABC′^ are equilateral in the same orien- tation. Prove that — with what one exception? — the line segments AA′, BB′, CC′^ have the same length, and make 60◦^ angles with each other (extended if necessary).

[It is also known that these three lines are concurrent; when the equilateral triangles are external to 4 ABC, the point of intersection is known as the Fermat point of 4 ABC — yes, the same Fermat that Problem 1 should bring to mind. I do not ask that you prove the concurrence, which cannot be easily obtained using the arithmetic of complex numbers.]

  1. Let a 6 = 1 be a positive real number. Show that the image of the unit circle (the set of complex numbers z = x + iy such that x^2 + y^2 = 1) under the function z 7 → z + a/z is an ellipse. What happens when a = 1, or a is a complex number?
  2. Show that the area A enclosed by a simple closed curve C in the complex plane is given by the contour integral

A =

2 i

C

¯z dz.

(An integral

f dz over some path in the complex plane is interpreted as the line integral

f dx + i

f dy where x and y are the real and imaginary parts of z = x + iy — imagine that dz = dx + i dy. The integrand ¯z is the complex conjugate x − iy of z = x + iy. We’ll soon have a lot more to say about such integrals.)

This problem set is due Monday, September 22, at the beginning of class.