The value of $i\log\left(x-i\right)+i^2\pi+i^3\log\left(x+i\right)+i^4\left(2\tan^{-1}x\right)$, (where, $x>0$ and $i=\sqrt{-1}$), is :
Step-by-Step Solution
Key Concept: Separate the expression into real and imaginary parts by recognizing that $\log(x±i)$ produces complex values, then use the complementary angle identity $\tan^{-1}x+\tan^{-1}(1/x)=\pi/2$.
We simplify using powers of $i$: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$. The expression becomes $i\log(x-i)-\pi-i\log(x+i)+2\tan^{-1}x$. Grouping real and imaginary parts: Real part is $-\pi+2\tan^{-1}x$, and imaginary part is $i[\log(x-i)-\log(x+i)]=i\log\frac{x-i}{x+i}$. For $x>0$, we have $\frac{x-i}{x+i}=\frac{(x-i)^2}{x^2+1}=\frac{x^2-1-2xi}{x^2+1}$, which gives $\log\frac{x-i}{x+i}=-2i\tan^{-1}(1/x)$. Therefore the imaginary part becomes $i(-2i\tan^{-1}(1/x))=2\tan^{-1}(1/x)$. Using $\tan^{-1}x+\tan^{-1}(1/x)=\pi/2$ for $x>0$, the real part is $-\pi+2(\pi/2-\tan^{-1}(1/x))=0$, and the imaginary part is also $0$.
Correct Answer: 1