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) = xso 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.