If -p-1 r e ip -1 where - denotes the continued product- then the most general value of - isA- 2 n -rr-1- n - ZB- 2 n -rr-1 n - ZC- 4 n -rr-1- n - ZD- 4 n -rr-1- n - Z

Given, ∏_p = 1^re^i p θ= 1 Hence, e^i θ· e^2i θ⋯ e^ri θ = 1 ⇒ e^ir(r + 1) θ2 = e^i2nπ,n∈ℤ ⇒θ = 4nπr(r + 1),n∈ℤ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Euler's Representation

If ∏p=1 r e i...

Question

If $r \prod p = 1 e^{i p θ} = 1$ where $\prod$ denotes the continued product, then the most general value of $θ$ is

\frac{2 n π}{r (r - 1)}, n \in Z

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 n π}{r (r + 1)} n \in Z

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4 n π}{r (r - 1)}, n \in Z

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4 n π}{r (r + 1)}, n \in Z

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

7 \cos^{2} θ + 3 \sin^{2} θ = 4

θ = 2 n π \pm \frac{π}{6}, n \in Z

θ = 2 n π \pm \frac{2 π}{3}, n \in Z

θ = n π \pm \frac{π}{3}, n \in Z

4 \sin^{2} θ = 1

2 n π \pm \frac{π}{3}, n \in Z

n π \pm \frac{π}{3}, n \in Z

n π \pm \frac{π}{6}, n \in Z

2 n π \pm \frac{π}{6}, n \in Z

\sqrt{3} sec 3 x = 2

4 \sin^{2} x = 1

2 n π \pm \frac{π}{3}, n \in Z

n π \pm \frac{π}{3}, n \in Z

n π \pm \frac{π}{6}, n \in Z

2 n π \pm \frac{π}{6}, n \in Z

7 \cos^{2} x + 3 \sin^{2} x = 4

x = 2 n π \pm \frac{π}{6}, n \in Z

x = 2 n π \pm \frac{2 π}{3}, n \in Z

x = n π \pm \frac{π}{3}, n \in Z

Join BYJU'S Learning Program