Question

If $z=\frac{(1+i)(1+2i)(1+3i)\dots \dots \dots (1+ni)}{(1-i)(2-i)(3-i)\dots \dots \dots (n-i)}$, where $i=\sqrt{-1},n\in N$, then principal argument of $z$ can be -

• A. $0$
• B. $\frac{\pi }{2}$
• C. $-\frac{\pi }{2}$
• D. $\pi$

Answer 1

Answer: (A), (B), (C), (D)

$\pi$

$z=\frac{(i-i^2)(2i-i^2)\dots \dots \dots (ni-i^2)}{(1-i)(2-i)\dots \dots (n-i)}=i^n$

$z$ can be $1,-1,i,-i$

So, $\arg z$ can be $0,\pi ,\frac{\pi }{2},-\frac{\pi }{2}$