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Assignment 1 Solutions

Page 5 #1

These are straightforward applications of the formulas for sums, differences, products, and quotients. I won't write them all out. Here is part (d) as an example:

   1 + 2i   2 - i     (1+2i)(3+4i)    (2-i)(-5i)
   ------ + -----  =  ------------  + ----------
   3 - 4i    5i       (3-4i)(3+4i)    (5i)(-5i)
   
   
                      [(1)(3)-(2)(4)] + [(1)(4)+(2)(3)]i   -5 - 10i
                   =  ---------------------------------- + ---------
                           2      2                            2
                          3   +  4                            5
   
                      -5 + 10i      -5 - 10i
                   =  --------   +  --------
                          25           25
   
                     -10 + 0i     -2
                   = --------  =  ---
                        25         5
   

Page 5 #4 (6th edition: #12)

In ordered pair notation, this equation becomes (x^2-y^2, 2xy) + (x,y) + (1,0) = (0,0); in other words, (x^2-y^2 + x + 1, 2xy + y) = (0,0). (In normal complex number notation we'd write this as (x^2-y^2+x+1) + (2xy+y)i = 0).

This means we must solve the simultaneous equations x^2-y^2+x+1=0 and 2xy+y=0. The second equation factors as y(2x+1)=0 which means either y=0 or 2x+1=0. In the case y=0, the first equation becomes x^2+x+1=0 which has no real solutions (the discriminant b^2-4ac = 1^2 - (4)(1)(1) = -3 < 0), so the only possible case is the second case 2x+1=0. In this case x=-1/2; plugging this into the first equation gives (1/4) - y^2 - 1/2 + 1 = 0, or 3 - 4y^2 = 0. The solution to this is y = +/- sqrt(3)/2.

Therefore, since x=-1/2 and y = +/- sqrt(3)/2, the two complex solutions to the original equation are z = ( -1/2, sqrt(3)/2) and z = (-1/2, -sqrt(3)/2 ).

Page 6 #6 (6th edition: #8(a))

From the definition of complex number addition, if we have three complex numbers z1= (x1,y1), z2=(x2,y2), and z3=(x3,y3), then (z1+z2)+z3= (x1+x2,y1+y2)+(x3,y3) = ((x1+x2)+x3, (y1+y2)+y3). Since addition of real numbers is associative, this equals (x1+(x2+x3),y1+(y2+y3)) = (x1,y1)+(x2+x3,y2+y3) = z1+ (z2+z3) (using the definition of complex number addition again).

This proves that (z1+z2)+z3= z1+(z2+z3).

Page 6 #10 (6th edition: #4)

Suppose z is the pair (x,y). The product iz means (0,1)(x,y) = ((0)(x)-(1)(y),(0)(y)+(1)(x)) = (-y,x). Therefore, Re(iz) = -y = -Im(z) and Im(iz) = x = Re(z). This proves parts (a) and (b).

For part (c), recall that 1/(x+iy) = (x-iy)/(x^2+y^2) for any complex number x+iy. If z=x+iy, this means 1/z = (x-iy)/(x^2+y^2) = u + iv where u=x/(x^2+y^2) and v=(-y)/(x^2+iy^2).

Now applying the same formula to the complex number 1/z = u + iv, we get 1/(1/z) = (u - iv)/(u^2+v^2). Calculating u^2+v^2 gives 1/(x^2+y^2), so 1/(1/z) = (u-iv)(x^2+y^2) = [x/(x^2+y^2) - i(-y)/(x^2+y^2)](x^2+y^2) = x + i y = z.

For part (d), recall that if z = (x,y) then (-1)z means the product (-1,0)(x,y) = ((-1)(x)-(0)(y), (-1)(y)+(0)(x)) = (-x,-y) = -z.

Page 6 #15 (6th edition: #14)

   z1 z2                1
   -----  =  (z1 z2)  -----   (by equation (13) in the book (5th edition))
   z3 z4              z3 z4
   
                      1    1
          =  (z1 z2) --   --  (by equation (14))
                     z3   z4
   
                   1         1
          =  ( z1 -- )  (z2 -- )   (by associativity of multiplication)
                  z3        z4
   
             z1 z2
          =  -- --  (by equation (13) again)
             z3 z4
   

The cancellation law follows from this by writing (z z_1)/(z z_2) = (z/z) (z_1/z_2) = (z_1/z_2).

Page 11 #1

For part (a):

(PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY)

The other parts are similar. Note that for part (d), even though x_1 and y_1 are unspecified, you know that z_1+z_2 = 2x_1 is purely real (hence a horizontal vector) and z_1-z_2 = 2iy_1 is purely imaginary (hence a horizontal vector). In the case where x_1, y_1 > 0, the picture looks as follows:

(PICTURE AVAILABLE IN GRAPHICAL VERSION ONLY)

Page 11 #2

Part (a) follows because

   ______     _
   _          _   __
   z + 3i  =  z + 3i  = z + (-3i)  = z - 3i

(the first inequality is from the law

   _______   __   __
   z1 + z2 = z1 + z2
, the second from the law that
   _
   _
   z = z
(and the fact that the conjugate of 3i=0+3i is 0-3i, i.e. -3i).

Parts (b) and (c) are similar; I won't write them all out.

Part (d) follows because

      _                     _
   |(2z+5)(sqrt 2 - i)| = |2z+5| |sqrt 2 - i|  (since |z1z2| = |z1| |z2|)
   
               _
   = sqrt(3) |2z + 5|   (since |sqrt(2)-i| = sqrt(3))
   
             _______
               _                      _
   = sqrt(3) |2z + 5|   (since |w| = |w| for all w)
   
              __
               _   _           _____   __   __
   = sqrt(3) |2z + 5 |  (since z1+z2 = z1 + z2 )
   
                _
              _ _   _          ____   __ __
   = sqrt(3) |2 z + 5 | (since z1z2 = z1 z2 )
                                                 _
                               _      _          _
   = sqrt(3) |2z + 5|   (since 2 = 2, 5 = 5, and z = z)

Page 11 #4

First we prove that 2ab <= a^2 + b^2 for all real numbers a and b. We do this by observing that 0 <=(a-b)^2 = a^2 + b^2 - 2ab and the desired inequality follows by adding 2ab to each side.

Now,

                    2          2         2
   (|Re z| + |Im z|)  =  (Re z)  + (Im z)  + 2 |Re z| |Im z|
   
                               2         2          2          2
                      <= (Re z)  + (Im z)  +  (Re z)   + (Im z)
   
                                   2         2
                       = 2 ( (Re z)  + (Im z) )
   
                              2
                       = 2 |z|

Since both | Re z| + |Im z | and | z | are non-negative, it is legitimate to take the square roots of both sides to conclude that |Rez| + |Imz| <= sqrt(2) |z|, as desired. (For non-negative numbers A and B, A <= B if and only if A^2 <= B^2).

Page 11 #6 (6th edition: #14)

For part (a), note that z is real if and only if Im z = 0, i.e.,

       _
   z - z
   ----- = 0
     2i
, which is true if and only if
       _                   _
   z - z = 0    (i.e., z = z)

For part (b): if z is either real or pure imaginary, then either

   _        _                            _ 2   2
   z = z or z = -z,  and in either case (z) = z .
   
                   _ 2   2            2    _ 2        _      _
   Conversely, if (z) = z , then 0 = z  - (z)  = (z - z)(z + z)
             _            _                    _        _
   so either z - z = 0 or z + z = 0, so either z = z or z = -z,
so z is either real or pure imaginary.

Page 11 #10 (6th edition: #9)

   One of the variant forms of the triangle inequality gives us
   |z3 + z4| >=  | |z3| - |z3| |, and therefore
   
        1                   1
    ---------   <=  --------------- 
    |z3 + z4|       | |z3| - |z4| |
   
   (provided the denominators are both positive, which is guaranteed
   by |z3| being not equal to |z4|)
   
   The triangle inequality also gives
   
   |z1 + z2| <= |z1| + |z2|.
   
   Therefore, 
   
     | z1 + z2 |   |z1+z2|              1
     | ------- | = ------- = |z1+z2| --------
     | z3 + z4 |   |z3+z4|           |z3+z4|
   
                                 1
            <=  (|z1| + |z2|) ---------------
                              | |z3| - |z4| |
   
                 |z1| + |z2|
             = --------------- .
               | |z3| - |z4| |
   

Page 11 #11 (6th edition: #10)

(a) Circle of radius 1 centred at 1-i.

(b) Closed disk (circle together with the points inside the circle) of radius 3 centred at -i.

(c) If z=x+iy,

      _
   Re(z - i)  =  Re (x - iy - i) = Re (x - (y+1)i) = x
so the sketch consists of the line x=2.

(d) Since |2z-i|=4 is equivalent to |z-(1/2)i| = 2, the sketch is a circle of radius 2 centred at 0.5i.

Page 11 #13 (6th edition: #12)

     4     2           2       2           2        2
   |z  - 4z  + 3| = |(z  - 1)(z  - 3)| = |z  - 1| |z  - 3|
   
                                2                2
                        >=  | |z | - |1| |   | |z | - |3| |
   
                                 2            2
                         =  | |z| - 1 |  | |z|  - 3 |
                                                          2    2
                         =  (4-1)(4-3) = 3      (since |z|  = 2  = 4)
   
   
   Therefore,
   
   |     1      |         1             1
   |------------| = -------------  <=  ---
   | 4     2    |     2     2           3
   |z  - 4z  + 3|   |z  - 4z  +3|
   

Page 12 #17

Since two non-negative numbers are equal if and only if their squares are equal, the equation can be rewritten

    2            2           ____           _ __     _    __   _      __
   R  =  |z - z0|  = (z-z0) (z-z0) = (z-z0)(z-z0) = zz - zz0 - zz0 +z0z0
   
                              ___
              2      2   __    __
         = |z| + |z0| + zz0 + zz0 
   
              2      2        __
         = |z| + |z0| + 2 Re(zz0)
   

Page 12 #18

                                    _ 2        _ 2
    2   2         2        2   (z + z)    (z - z)
   x - y =  (Re z) - (Im z)  = -------  - -------
                                    2          2
                                   2       (2i)
   
               _ 2      _ 2
            (z+z)    (z-z)
          = -----  + -----
              4        4
   
             2      _   _ 2    2       _    _ 2
            z + 2 z z + z   + z  - 2 z z  + z
          = -----------------------------------
                          4
   
             2   _ 2
            z  + z
          = --------
               2
   
       2    2                      2  _ 2
   so x  - y  = 1  if and only if z + z  = 2.

Page 12 #19

(a) This equation represents the set of points for which the sum of the distance to 4i (i.e., the point (0,4)) and the distance to -4i (i.e., the point (0,-4)) is 10. Geometrically, the set of points for which the sum of the distance to point P and the distance to point Q is a fixed value is an ellipse with foci at P and Q.

(b) This equation represents the set of points that are equidistant from 1 (i.e., the point (1,0)) and -i (i.e., the point (0,-1)). This set is the line bisecting the line segment joining these two points, which is the line through the origin with slope -1.



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