1. Show that (a) (1+i)'= exp P(+2x) exp (112) 1 (b) i2i (- 4 = exp[(4n+1)л] (n = 0, ±1, ±2,...). (n = 0, ±1, ±2,...); Section 37 la) Show that (1+i)'= exp(- +2 nπT) exp (i (2) (0 (+7322) Proof (1+;) in Polar form: r = √12²+12 = √2 0 = arctum (+) = tun" " (1) = HT (1+i) = √√2 (cos() + is in (1) (n=12122) (1+i) in exponentul form using Euler's formula? (1+i) = √√2 et #) (1+i); = eiloy (√2e) ei i log (√2 e₁ =) = ; (Inll√zei =)) + Arg (√2017)) = illn (√2) +1 (4+2πh)) = ihn (12) - (+2πh) (4 So (1+i)' = @ilm (√2) - ( + 2πh) = V e - (+2πn) in (√2) e e- (= +2πth) eien (2)
1. Show that (a) (1+i)'= exp P(+2x) exp (112) 1 (b) i2i (- 4 = exp[(4n+1)л] (n = 0, ±1, ±2,...). (n = 0, ±1, ±2,...); Section 37 la) Show that (1+i)'= exp(- +2 nπT) exp (i (2) (0 (+7322) Proof (1+;) in Polar form: r = √12²+12 = √2 0 = arctum (+) = tun" " (1) = HT (1+i) = √√2 (cos() + is in (1) (n=12122) (1+i) in exponentul form using Euler's formula? (1+i) = √√2 et #) (1+i); = eiloy (√2e) ei i log (√2 e₁ =) = ; (Inll√zei =)) + Arg (√2017)) = illn (√2) +1 (4+2πh)) = ihn (12) - (+2πh) (4 So (1+i)' = @ilm (√2) - ( + 2πh) = V e - (+2πn) in (√2) e e- (= +2πth) eien (2)
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
Can you correct what I did wrong in part a?
![1. Show that
(a) (1+i)'= exp P(+2x) exp (112)
1
(b)
i2i
(-
4
= exp[(4n+1)л] (n = 0, ±1, ±2,...).
(n = 0, ±1, ±2,...);](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6389447-1237-4af0-b5c6-eb1260425b55%2Fd34475c5-a202-47a7-be16-0d03eed67be7%2Fveskg2n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. Show that
(a) (1+i)'= exp P(+2x) exp (112)
1
(b)
i2i
(-
4
= exp[(4n+1)л] (n = 0, ±1, ±2,...).
(n = 0, ±1, ±2,...);
Transcribed Image Text:Section 37
la) Show that (1+i)'= exp(- +2 nπT) exp (i (2) (0 (+7322)
Proof
(1+;) in Polar form:
r = √12²+12 = √2
0 = arctum (+) = tun" " (1)
=
HT
(1+i) = √√2 (cos() + is in (1)
(n=12122)
(1+i) in exponentul form using Euler's formula?
(1+i) = √√2 et
#)
(1+i); = eiloy (√2e)
ei
i log (√2 e₁ =) = ; (Inll√zei =)) + Arg (√2017))
= illn (√2) +1 (4+2πh))
= ihn (12) - (+2πh)
(4
So (1+i)' = @ilm (√2) - ( + 2πh)
=
V
e
- (+2πn) in (√2)
e
e- (= +2πth) eien (2)
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