Download Course: Engineering Mathematics I (MATH 1180) Group I: CIV

Course: Engineering Mathematics I (MATH 1180)
Assignment #1: COMPLEX NUMBERS
MATH 1180: 
Group I: CIV, ELEC, SURV.
Assignment #1: COMPLEX NUMBERS I
Lecturer: Mr. Oral Robertson
Date: September 29, 2010
Due on: Monday, October 4, 2010.
1. (a) Prove using the “conjugate method”: |z1 + z2 + z3|  |z1| + |z2| + |z3|.
 Hint: Consider |z1 + z2 + z3|2 , and that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca .
(b) By using the polar form of two general complex numbers z1 and z2 show that:
arg(z1 z2) = arg(z1) + arg(z2) and arg(z1/z2) = arg(z1) – arg(z2)
(c) Starting with the standard inequality result |z1 + z2|  |z1| + |z2|, prove that |z1 – z2|  |z1| – |z2|.
 Hint: Replace z1 by (z1 – z2) without loss of generality.
2. By considering the individual modulus and argument of each bracketed term, use the rules of complex
numbers (products, quotients, and powers) to express the following complex numbers in polar form.
(a)
 1  i 3 1  i 
 3  i
1  i 3  1  i
(b)
 3  i
3
1  i 3   1  i 
5
4
(c)
5
6
 2  2i 4
3. (a) Show that the locus of a complex number z = x + yi moving so that it is equidistant from two fixed
points represented by the complex numbers z1 = a + bi and z2 = c + di is a straight line that is the
perpendicular bisector of the line segment connecting z1 and z2 .
(b) Find the locus of the complex number z defined by |2z + 1| = |z – 1|. Describe fully its geometry.
(c) Show that the locus of a complex number z defined by |z – 1| + |z + 1| = 4 is an ellipse having
equation:
y2
x2

 1 . Express in words, the motion of z based on the initial modulus-based equation given.
4
3
 zi  π
(d) Find the equation of the locus defined by : arg
  . Describe fully its geometry.
 z  2 4
4. Obtain the following transformations (T) results from the xy-plane onto the uv-plane, where z = x + iy
and w = u + iv:
(a) Map the straight line y = 2x + 1 under T: w = 2z – 1, to get the image v = 2u + 4 .
1
z
(b) Map the circle x2 + y2 = 1 under T: w  z  , to get the image v = 0 (a horizontal line through O).
1
z
(c) Map the straight line y = 2x – 1 under T: w  , to get the image u2 – 2u + v2 – v = 0 (a circle).
5. By substituting  = ix, into the following trigonometric identities, derive their analogous hyperbolic
forms:
(a) cos2 + sin2 = 1
(b) sec2 = 1 + tan2
(c) sin2 = 2 sin cos
(d) cos2 = cos2 – sin2 
6. (a) Use the results sin x 
e ix  e ix
e ix  e ix
and cos x 
to show that: sin3 cos2 = ½ (sin5 + sin).
2i
2
(b) Use the same technique in part (a) to obtain the more general identity:
sinA cosB = ½ [sin(A+B) + sin(A – B)].
7. For z = x + iy, obtain the following in the form a + ib, hence calculate their moduli and arguments:
(i) (a) sin z
(b) sin z
(c) tan z
(d) cos z
(e) cos z
(f) cot z
(ii) It is seen that sin z = sin z , and cos z = cos z , use this fact to prove that: tan z = tan z .
 Hint: Recall that the quotient of the conjugates equals the conjugate of the quotient.
  sinh 2 y 
.
 sin 2 x 
(iii) Also show that: argcot z   tan 1 
1