How prove this $arg((1+ia)(2+ia)(3+ia)\cdots(n+ia))=\arctan{\frac{a}{1}}+\arctan{\frac{a}{2}}+\cdots+\arctan{\frac{a}{n}}$

Question

let $i^2=-1,a>0$, show that $$arg((1+ia)(2+ia)(3+ia)\cdots(n+ia))=\arctan{\dfrac{a}{1}}+\arctan{\dfrac{a}{2}}+\cdots+\arctan{\dfrac{a}{n}}$$

I can't How prove this equation,Thank you

because $n=1$.we have $$(1+ia)=\arctan{a}\Longleftrightarrow \tan{(1+ia)}=a?$$ then I can't.Thank you

i think the question should be to prove $$arg((1+ai)(2+ai)...(n+ai))=\sum_{k=1}^{n}\arctan(\frac{a}{k})$$ they might be discussing about the arguement. — Suraj M S, Feb 28 '14 at 17:26
The main point is : "the complex argument of a product of two numbers is equal to the sum of their arguments, "http://mathworld.wolfram.com/ComplexArgument.html. — lab bhattacharjee, Feb 28 '14 at 17:48

score 0 · Answer 1 · edited Feb 28 '14 at 18:57

0

let $$k+ia=r_ke^{i\theta_k}=r_k(\cos \theta_k+i\sin \theta_k)$$ where $\theta_k=\arctan(\frac{a}{k})$ is the arguement of the term. now what do you get in the arguement part when you multiply all the terms. you get a $r_1r_2\cdots r_ne^{i(\theta_1+\theta_2+\cdots+\theta_n)}$

edited Feb 28 '14 at 18:57

Michael Hardy

1

answered Feb 28 '14 at 17:37

Suraj M S

1,891

How prove this $arg((1+ia)(2+ia)(3+ia)\cdots(n+ia))=\arctan{\frac{a}{1}}+\arctan{\frac{a}{2}}+\cdots+\arctan{\frac{a}{n}}$

1 Answers1