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This is the second of our many quizzes on De Moivre's theorem. Click on this to access more such competitive quizzes on Number system.

De Moivre's theorem Quiz 2

mathematics

 What is the polar form of the equation $\left( s+ig \right)$. Where, $s$ and $r$ are integers.

 Though it is the standard formula for the polar form of a complex number, find the value of $\theta $ based on the basic formula. 

  It is the standard formula from De moivre’s theorem that any complex number which is in the form of $\left( s+ig \right)$ can be written as ${{r}^{n}}\left( \cos \theta +i\sin \theta  \right)$ in polar form. $ \\ $ Where, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{g}{s}$.

 Solve ${{\left( a+ib \right)}^{n}}-{{\left( a-ib \right)}^{n}}$.

 Convert the given equation into polar form and then use De moivre’s theorem to simplify the equation.

  Given ${{\left( a+ib \right)}^{n}}+{{\left( a-ib \right)}^{n}}$. $ \\ $ Solve the equation by using the polar form. $ \\ $ Polar form of any complex number $p+iq$ will be $r\left( \cos \theta +i\text{sin}\theta  \right)$. Whereas $r=\sqrt{{{p}^{2}}+{{q}^{2}}}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$. $ \\ $ $\Rightarrow {{r}^{n}}{{\left( \cos \alpha +i\sin \alpha  \right)}^{n}}-{{r}^{n}}{{\left( \cos \alpha -i\sin \alpha  \right)}^{n}}$ $ \\ $ From De moivre’s theorem $ \\ \Rightarrow {{r}^{n}}\left( \cos n\alpha +i\sin n\alpha  \right)-{{r}^{n}}\left( \cos n\alpha -i\sin n\alpha  \right) \\ \Rightarrow {{r}^{n}}\left( 2i\sin n\alpha  \right) \\ $ Hence the correct answer is $2{{r}^{n}}i\sin n\alpha $ where $r=\sqrt{{{a}^{2}}+{{b}^{2}}}$ and $\alpha ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$.

 If $x=cis\theta $. Then find the value of ${{x}^{n}}-\dfrac{1}{{{x}^{n}}}$? 

 Raise the given equation to power $n$ and reciprocal it to get the value of $\dfrac{1}{{{x}^{n}}}$. Then subtract the equation of $\dfrac{1}{{{x}^{n}}}$ from the equation of ${{x}^{n}}$ to obtain the result.

$\text{As it is given that }x=cis\theta . \\ \text{Now raising the power to }n\text{ on both sides} \\ \Rightarrow {{x}^{n}}={{\left( cis\theta  \right)}^{n}}=cisn\theta .................\left( 1 \right) \\ \text{Hence from the formula }{{\left( cis\theta  \right)}^{n}}=cis\left( n\theta  \right) \\ \text{Reciprocal the above equation} \\ \Rightarrow \dfrac{1}{{{x}^{n}}}=\dfrac{1}{cis\left( n\theta  \right)} \\ \text{From the formula }\dfrac{1}{cis\theta }=cis\left( -\theta  \right) \\ \Rightarrow \dfrac{1}{{{x}^{n}}}=cis\left( -n\theta  \right).............\left( 2 \right) \\ \text{Subtracting the (2) from (1) we get,} \\ {{x}^{n}}-\dfrac{1}{{{x}^{n}}}=2i\sin n\theta  $

 Find the imaginary part of ${{e}^{{{e}^{i\theta }}}}$.

 From Euler’s theorem we can write ${{e}^{i\theta }}=\cos \theta +i\sin \theta $. So, therefore use this theorem and then simplify the equation to get the answer.

 Given to solve ${{e}^{{{e}^{i\theta }}}}$. $ \\ $ As ${{e}^{i\theta }}$ states that ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ $ \\ $ Now, ${{e}^{{{e}^{i\theta }}}}={{e}^{\cos \theta +i\sin \theta }}$ So, ${{e}^{{{e}^{i\theta }}}}={{e}^{\cos \theta +i\sin \theta }}$ can be written as $ \\ {{e}^{{{e}^{i\theta }}}}={{e}^{\cos \theta }}\cdot{{e}^{i\sin \theta }} \\ $ Therefore, the imaginary part for the given equation will be ${{e}^{i\sin \theta }}$.

 Solve $29{{\left( 1+\cos \alpha +i\sin \alpha  \right)}^{n}}+41{{\left( 1+\cos \alpha -i\sin \alpha  \right)}^{n}}$.

 Expand $1+\cos \alpha $ and $\sin \alpha $ in the given equation and then simplify the equation to apply the De moivre’s theorem. After applying the De moivre’s theorem, solve the equation to get the solution.

 Given $29{{\left( 1+\cos \alpha +i\sin \alpha  \right)}^{n}}+41{{\left( 1+\cos \alpha -i\sin \alpha  \right)}^{n}}$ In the given equation $1+\cos \alpha $ and $\sin \alpha $ can be written as $2{{\cos }^{2}}\dfrac{\alpha }{2}$ and $2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}$. Substituting the formulas and rewriting the equation.  $ \\ 29{{\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}+i2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} \right)}^{n}}+41{{\left( 2{{\cos }^{2}}\dfrac{\alpha }{2}-i2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2} \right)}^{n}} \\ $ Taking $2\cos \dfrac{\alpha }{2}$ common from both the terms $ \\ {{29.2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right){{\left( \cos \dfrac{\alpha }{2}+i\sin \dfrac{\alpha }{2} \right)}^{n}}+{{41.2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right){{\left( \cos \dfrac{\alpha }{2}-i\sin \dfrac{\alpha }{2} \right)}^{n}} \\ $ Apply De moivre’s theorem $ \\ {{29.2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right)\left( \cos n\dfrac{\alpha }{2}+i\sin n\dfrac{\alpha }{2} \right)+{{41.2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right)\left( \cos n\dfrac{\alpha }{2}-i\sin n\dfrac{\alpha }{2} \right) \\ \Rightarrow {{2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right)\left( 29\left( \cos n\dfrac{\alpha }{2}+i\sin n\dfrac{\alpha }{2} \right)+41\left( \cos n\dfrac{\alpha }{2}-i\sin n\dfrac{\alpha }{2} \right) \right) \\ \Rightarrow {{2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right)\left( 29\cos n\dfrac{\alpha }{2}+29i\sin n\dfrac{\alpha }{2}+41\cos n\dfrac{\alpha }{2}-41i\sin n\dfrac{\alpha }{2} \right) \\ \Rightarrow {{2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right)\left( 70\cos n\dfrac{\alpha }{2}-12i\sin n\dfrac{\alpha }{2} \right) \\ $ Therefore ${{2}^{n}}\cdot{{\\cos }^{n}}\left( \dfrac{\alpha }{2} \right)\left( 70\cos n\dfrac{\alpha }{2}-12i\sin n\dfrac{\alpha }{2} \right)$ is the solution.