(k) (zw)-1 = w-!z-1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please do question 22 part k and please show each step
![Proposition 4.2.21. Given z and w are complex numbers, then z+ w =
z+ w.
PROOF. We may write z as a + bi and w as c+ di. Then
Z+ w = a + bi +c+di
= (a – bi) + (c – di)
= (a + c) – (b+ d)i
= (a + c) + (b+ d)i
by definition of conjugate
commutative, associative
by definition of conjugate
%3D
= z + w
by definition of complex addition
Exercise 4.2.22. Prove each of the following propositions (follow the style
of Proposition 4.2.21).
(a) (z) = z
(g) |z|3 = |2°| (*Hint*)
(b) z. w = zW
(h) z-1
(*Hint*)
|z|2
(c) If a is real, then az = az
(d) |z| = |7|
(i) ]z-1| :
1
(*Hint*)
|z|
(e) zz = |z|2
(j) (z)-1 = =1
(f) |zw| = |z||w|
(k) (zw)-1 = w-1z-1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2F4a1a1f2d-929e-4fa2-ae23-073473e9be58%2Fjmmyete_processed.png&w=3840&q=75)
Transcribed Image Text:Proposition 4.2.21. Given z and w are complex numbers, then z+ w =
z+ w.
PROOF. We may write z as a + bi and w as c+ di. Then
Z+ w = a + bi +c+di
= (a – bi) + (c – di)
= (a + c) – (b+ d)i
= (a + c) + (b+ d)i
by definition of conjugate
commutative, associative
by definition of conjugate
%3D
= z + w
by definition of complex addition
Exercise 4.2.22. Prove each of the following propositions (follow the style
of Proposition 4.2.21).
(a) (z) = z
(g) |z|3 = |2°| (*Hint*)
(b) z. w = zW
(h) z-1
(*Hint*)
|z|2
(c) If a is real, then az = az
(d) |z| = |7|
(i) ]z-1| :
1
(*Hint*)
|z|
(e) zz = |z|2
(j) (z)-1 = =1
(f) |zw| = |z||w|
(k) (zw)-1 = w-1z-1
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