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### $\frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} = z$${\mathop{\rm Im}\nolimits} (z) = 0$$\theta = ?$

$z = \frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} \times \frac{{1 - i\cos \theta }}{{1 - i\cos \theta }}$

$= \frac{{1 - i(\cos \theta + \sin \theta ) + {i^2}\sin \theta \cos \theta }}{{1 - {i^2}{{\cos }^2}\theta }}$

$= \frac{{1 - \sin \theta \cos \theta - i(\cos \theta + \sin \theta )}}{{1 + {{\cos }^2}\theta }}$

$\because {\mathop{\rm Im}\nolimits} (z) = 0,\cos \theta + \sin \theta = 0$

$\Rightarrow \tan \theta = -1$

$\Rightarrow \theta = n\pi - \frac {\pi}{4}$,$n\in I$

### $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, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$