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$i^i$ where $i=\sqrt {-1}$ is a real number. T/F

We have, $e^{i\theta}=cos\theta +i sin\theta $

Let, $\theta = 2n\pi + \frac {\pi}{2}$

Then, $e^{i.(2n\pi + \frac {\pi}{2})}=cos (2n\pi + \frac {\pi}{2}) +i sin (2n\pi + \frac {\pi}{2}) $

$\Rightarrow e^{i.(4n+1) \frac {\pi}{2}}=0+i.1 $

$\Rightarrow i.(4n+1)\frac {\pi}{2} = \ln i $

Let, $z=i^i$

$\Rightarrow \ln z = i.\ln i$

So, $\ln z = i.i(4n+1)\frac {\pi}{2}= i^2.(4n+1)\frac {\pi}{2} = -(4n+1)\frac {\pi}{2}$

$\Rightarrow z = e^{-(4n+1)\frac {\pi}{2}}$ = Real number

Hence, True.

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$