Unformatted text preview: inequality are nonnegative, so (x + u)2 + (y + v )2 ≤
2 2 2 2 2 x2 + y 2 +
2 2 u2 + v 2
2 (27) (x2 + y 2 ) (u2 + v 2 ) (28) (29) ←→ x + u + 2xu + y + v + 2yv ≤ x + y + u + v + 2 ←→ xu + yv ≤ (x2 + y 2 ) (u2 + v 2 ) Here we are no longer assured that both sides are nonnegative, since the left side can be less than zero. If it is, then the inequality holds, otherwise we need to investigate further. Therefore, assuming that the left side is nonnegative, we again square both sides: ←→ x2 u2 + y 2 v 2 + 2xuyv ≤ x2 u2 + x2 v 2 + y 2 u2 + y 2 v 2 ←→ 0 ≤ x v + y u − 2xuyv ←→ 0 ≤ (xv − yu) ←→ 0 ≤ xv − yu (always true) That is, z + w ≤ z  + w if and only if 0 ≤ xv − yu , so z + w ≤ z  + w holds. Note: xv − yu =  (¯w) =  (z w) z ¯ b) First, we rewrite the triangle inequality: a + b − a ≤ b Therefore, a + b ≡ z and a ≡ w −→ z  − w ≤ z − w (see Figure 3 for interpretation) a + b ≡ z and a ≡ −w −→ z  − w ≤ z + w (see Figure 3 for interpretation) (35) (36) (34)
2 22 22 (30) (31) (32) (33) Problem 4 (10 points)
Find all the roots of sin z = −i. 5 z+w zw z w
Figure 3: We can associate vectors to the various complex quantities. Thus, z  − w ≤ z − w states that the diﬀerence in the lengths of z and w cannot be larger than the length of z − w; equality is reached when z and w are parallel (i.e., Arg(z ) = Arg(w)) and w ≤ z . z  − w ≤ z + w can be similarly seen (imagine the head of w connected to the tail of z ), with equalit...
View
Full
Document
This note was uploaded on 02/22/2010 for the course ACM 95A taught by Professor Nilesa.pierce during the Fall '06 term at Caltech.
 Fall '06
 NilesA.Pierce

Click to edit the document details