111 MAS FOUNDATION OF MATHEMATICS
ASSIGNMENT 12 HINTS
1. Multiply (a) (67i)(8+i), 2323 (b) ( + i)( i), 3232 1 1 (c) ( + i)(1 2i). 2 2
Hint: Apply the formula (a + bi)(c + di) = (ac bd) + (bc + ad)i. 2. Express z 1 in the form z 1 = a + bi for: (a) z = 6 + 8i, (b) z = 6 8i, (c) z = (6 + 7i)(8 i).
Hint: I only give z 1 for z = 6 + 8i. (6 + 8i)1 = 1 6 8i 6 8i 3 2 = = = i. 6 + 8i (6 + 8i)(6 8i) 100 50 25
Try the others by yourself. 3. Find (cos + i sin )1 . Hint: cos i sin 1 = = cos i sin . cos + i sin (cos i sin )(cos + i sin ) 4. Find the polar form for 1 2 (a) z = i, (b) z = 4i, 2 2 (d) z = 13 39 + i. 2 23
6 6 (c) z = + i 2 2
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Hint: I give solutions to (b) and (d) and leave others to you. (b) 4i = 4(cos + i sin ) (the modulus of 4i is 4 and the argument is 2 2 /2). 13 39 3 2 2 1 (d) + i = 13( + i) = 13(cos + i sin ) (the modu2 2 2 3 3 23 13 39 2 lus of + i is 13, and the argument is ). 2 3 23 5. Prove by direct multiplication that (cos( ) + i sin( ))2 = cos + i sin . 2 2 Hint: Do it by applying the multiplication rule. 6. Find the 6-th roots of the following complex numbers: (a) i, (b) 1 i, (c) 3, (d) 16 + 16 2i. Hint: I will give solutions to (a) and leave the others to you. (a) i = cos + i sin and the 6-th roots of i are cos( 2k + 2k ) + + i sin( ), 6 6
where k = 0, 1, 2, 3, 4, 5. That is, 6-th roots of i are 31 cos( ) + i sin( ) = + i; 6 6 2 2 3 3 cos( ) + i sin( ) = i; 6 6 31 5 5 cos( ) + i sin( ) = + i; 6 6 2 2 7 7 31 cos( ) + i sin( ) = i; 6 6 2 2 9 9 cos( ) + i sin( ) = i; 6 6 11 11 31 cos( ) + i sin( )= i; 6 6 2 2 7. If z is a complex number such that |z | = 1 (i.e. z z = 1), compute 2 2 |1 + z | + |1 z | . 2
Hint: Since z is given with |z | = 1, z is on the unit circle. Also note that |1 + z | is the distance between z and (1, 0) and |1 z | is the distance between z and (1, 0). Since the line connecting z and (1, 0) is orthogonal with the line connecting z and (1, 0), |1 + z |2 + |1 z |2 is equal to the square of the distance between (1, 0) and (1, 0), and hence equals to 4. You can also try to do like this: Let z = a + bi. Since |z | = 1, a2 + b2 = 1. Now |1 + z |2 + |1 z |2 = (1 + a)2 + b2 + (1 a)2 + b2 = 2 + 2(a2 + b2 ) = 4. 8. If x, y are complex, prove that ||x| |y || |x y |. Hint: Again, you can try to solve it by applying the geometric interpretation. |x| is the distance between x and the original; |y | is the distance between y and the original; |x y | is the distance between x and y . You can also try to let x = a + bi and c + di and then verify this inequality.
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