Download - Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos

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Page 1: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 2: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 3: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos

Assignment 6

Group 5: Apurv & Sanjay

Question

2 i. If sin(θ + iφ) = ρ(cosα + i sinα), prove that ρ2 = 1/2[cosh 2φ− cos 2θ]

Solution

ρ(cosα + i sinα) = sin (θ + iφ) (0.0.1)

= sin θ cos iφ+ cos θ sin iφ (0.0.2)

= sin θ coshφ+ i cos θ sinhφ (0.0.3)

For two complex numbers to be equal one of the necessary condition is that their magni-

tude must be equal. Then,

ρ2(cos2 α + sin2 α) = sin2 θ cosh2 φ+ cos2 θ sinh2 φ (0.0.4)

ρ2 = (1− cos2 θ) cosh2 φ+ cos2 θ(1− cosh2 φ) (0.0.5)

= cosh2 φ− cos2 θ (0.0.6)

=1 + cosh 2φ

2− 1 + cos 2θ

2(0.0.7)

=1

2[cosh 2φ− cos 2θ] Q.E.D (0.0.8)

1

Page 4: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 5: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 6: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 7: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 8: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 9: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 10: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 11: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos

ASSIGNMENT-6

Group-10(Bivin Geo George and Ritwik Mondal)

Question 7:

If

, then prove that

Answer:

Now,

Using the equation,

And similarly using

Therefore from previous,

Then take the conjugate of that,

Multiplying the two last equations we got that

Page 12: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos

That’s what it was in the question.

Page 13: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos
Page 14: Assignment 6 - prl.res.injayesh/solution6.pdf · Assignment 6 Group 5: Apurv & Sanjay Question 2 i. If sin( +i˚) = ˆ(cos +isin ), prove that ˆ2 = 1=2[cosh2˚ cos2 ] Solution ˆ(cos