Question

What is the argument of the complex number $\frac{(1+i)(2+i)}{3-i}$ where $i=\sqrt{-1}$ ?

• A. 0
• B. $\frac{\pi }{4}$
• C. $-\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

$\frac{(i+1)(2+i)}{(3-i)}$

Multiplying both denominator and numerator by $(3+i)$

$\Rightarrow \frac{(i+1)(2+i)(3+i)}{(3-i)(3+i)}=\frac{(3i+1)(3+i)}{10}=\frac{10i}{10}=i$

Argument of a complex number $x+iy$ is given by

$arg=\arctan \frac{y}{x}$

Then,$arg=\arctan \frac{1}{0}=\arctan \infty =\pi /2$