if cos(u+iv)=eiπ/6, u and v are real and v is positive, show that u=(2n-1)π/4 and v=loge(2cosπ/12)?
Sir we are getting u = nπ+π/4
but as per question we have to prove u=(2n-1)π/4
how to do this
- 6 months ago
Thank you Sir i will show my workingSource(s): w+1/w=√3+i ⇒w=[(√3+i)±√(2i√3-2)]/2 Let √(2i√3-2)=a+ib ⇒2i√3-2=a²-b²+i2ab Equating the real and imaginary parts a²-b²=-2–––(1) 2ab = 2√3–––(2) (a²+b²)²=(a²-b²)²+4a²b² =4+12=16 a²+b²=4–––(3) (1) + (3) ⇒ a=1 Hence b = √3 w=[√3+1+i(√3+1)]/2] w=e-v.cosu+i.e-v.sinu=a+ib tanu =b/a⇒u=tan-1(b/a)=tan-1(1) is my approach correct
- rotchmLv 76 months ago
Strange that no one has answered yet. I will therefore give you a walkthrough.
Let z = u+iv. Use cos(z) = (e^(iz) + e^(-iz))/2. Let w = e^(iz).
Also, eiπ/6 = √3/2 + i/2. Replacing all these into your original eqs leads to
w + 1/w = √3 + 1i
This is essentially a quadratic eqs & thus left for you to explicitly solve.
Thus, w = A + Bi where you found reals A & B.
Back to w = A + Bi = e^(iz) = e^(i(u+iv)) = e^(-v) * e^(iu) = e^(-v)*(cos(u) + isin(u)).
B = e^(-v)*sin(u)
A = e^(-v)*cos(u)
Dividing gives B/A = tan(u) thus u = your answer.
And adding the squares ...you get the idea now...
[The cos(π/12) is related to the cos(π/6) and the doubling formula].
To your update:
Yes. [but did you check the two solutions to the quadratic eqs?].
So here, A= B = (1+√3)/2. And for the other (perhaps) solution to the quadratic, you
will find that there too, A=B ...
Then... "tan-1(1)", yes. Thus gives π/4 + nπ, right?
And the sum of the squares...?