Question

If $a=1+i^2+i^4+\cdots +i^{2n}$ and $b={\cos }^{-1}\left(\left|\frac{1}{1+i}\right|\right)$, where $i=\sqrt{-1}$, then the value of $\tan \left(\frac{b}{a}\right)$ is

• A. $0$
• B. $1$
• C. cannot be determined
• D. depends on the value of $b$

Answer 1

The correct option is C cannot be determined

$a=1+i^2+i^4+\cdots +i^{2n}\Rightarrow a=1-1+1-1+\cdots +(-1{)}^n$
So, there are two values of $a$ depending on $n$.
When $n$ is odd, $a=0$
When $n$ is even, $a=1$

Now, $b={\cos }^{-1}\left(\left|\frac{1}{1+i}\right|\right)={\cos }^{-1}\left(\left|\frac{1-i}{2}\right|\right)$
$\Rightarrow b={\cos }^{-1}\frac{1}{\sqrt{2}}=\frac{\pi }{4}$

When $a=0$, $\tan \left(\frac{b}{a}\right)$ cannot be determined.
When $a=1$, $\tan \left(\frac{b}{a}\right)=1$
Hence, the value of $\tan \left(\frac{b}{a}\right)$ cannot be determined.