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This is the second of our many quizzes on graphing complex numbers. Click on this to access more such competitive quizzes on all topics of Complex numbers.

Graphing Complex Numbers Quiz 2

mathematics

 Convert the given complex number to polar form: ${{i}^{87}}$.

 Convert $z$ in the form of $a+ib$ by changing the power of $i$ in terms of 2 then find radius and the value of $\theta $ to represent complex number in the polar form as $z=r\left( \cos \theta +i\sin \theta  \right)$.

  $\begin{align} & \text{Given complex number: }z={{i}^{87}}\Rightarrow {{({{i}^{2}})}^{43}}.i=-i=0+(-1)i\,\,\,\,\,\,\,(\because {{i}^{2}}=-1)\\ & \text{ Polar form is }z=r(\cos \theta +i\sin \theta )\\ &\Rightarrow r=\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=1\,\,\,\,\,\,\,(\because a=0,b=-1)\\ & \text{Let }r\cos \theta =0\text{ and }r\sin \theta =-1\\ &\therefore \cos \theta =0\text{ and }\sin \theta =-1\\ & \text{So, polar form of }{{i}^{87}}\text{ is: }\cos \left( \dfrac{-\pi }{2} \right)+i\sin \left( \dfrac{-\pi }{2} \right) \end{align}$

 Convert the given complex number to polar form: ${{i}^{-35}}$.

 Convert z in the form of $a+ib$ by changing the power of $i$ in terms of 2 then find radius and the value of $\theta $ to represent complex number in the polar form as $z=r\left( \cos \theta +i\sin \theta  \right)$.

  $\begin{align} & \text{Given complex number: }z={{i}^{-35}}\Rightarrow \dfrac{1}{{{\left( {{i}^{2}} \right)}^{17}}.i}=\dfrac{i}{-{{i}^{2}}}=i=0+1(i)\\ & \text{Let }r\cos \theta =0\text{ and }r\sin \theta =1\\ & \text{On squaring on both sides and adding the equations, we obtain }\\ & \Rightarrow {{r}^{2}}=1\Rightarrow r=1\,\,\,\,\,\,\,\,\,(\text{Conventionally, }r>0 )\\ & \therefore \cos \theta =0\text{ and }\sin \theta =1\\ &\text{So, polar form of }{{i}^{-35}}\text{ is: }\cos \left( \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\pi }{2} \right) \end{align}$

 Find the value of $\dfrac{4{{i}^{668}}+2{{i}^{336}}+3{{i}^{266}}}{{{i}^{852}}-{{i}^{354}}}-\dfrac{{{i}^{236}}}{2{{i}^{146}}}$. 

 Change the power of $i$ in terms of 2 and 4, then put the value of ${{i}^{2}}\text{ and }{{i}^{4}}$ to simplify the expression. 

 $\begin{align} & \dfrac{4{{i}^{668}}+2{{i}^{336}}+3{{i}^{266}}}{{{i}^{852}}-{{i}^{354}}}-\dfrac{{{i}^{236}}}{2{{i}^{146}}} \\ & \Rightarrow \dfrac{4{{\left( {{i}^{4}} \right)}^{167}}+2{{\left( {{i}^{4}} \right)}^{84}}+3{{\left( {{i}^{2}} \right)}^{133}}}{{{\left( {{i}^{4}} \right)}^{213}}-{{\left( {{i}^{2}} \right)}^{177}}}-\dfrac{{{\left( {{i}^{4}} \right)}^{59}}}{2{{\left( {{i}^{2}} \right)}^{73}}}\,\,\,\,\,\,\,\,\,\left[ \because {{i}^{2}}=-1,\text{ }{{i}^{4}}=1 \right] \\  & \Rightarrow \dfrac{4+2-3}{1+1}+\dfrac{1}{2} \\  & \Rightarrow \dfrac{4}{2}=2  \end{align}$

 The argument of the complex number $\left( 5\sqrt{3}i-15{{i}^{2}} \right)\left( -12{{i}^{7}}-4\sqrt{3}{{i}^{10}} \right)$ is:

 Compare given complex number with the polar form i.e., $z=r\left( \cos \theta +i\sin \theta  \right)$ and find argument of ${{z}_{1}}\text{ and }{{z}_{2}}$. Now, sum both the arguments to find the required answer. 

 $\begin{align} & \text{We have, }{{z}_{1}}=5\sqrt{3}i-15{{i}^{2}}=15+5\sqrt{3}i\,\,\,\,\,\,\,\,\,(\because {{i}^{2}}=-1)\\ & \text{Let }{{r}_{1}}\cos \theta =15\text{ and }{{r}_{1}}\sin \theta =5\sqrt{3}.\\ & \text{On squaring both sides of the equations and adding: }\\ & \Rightarrow {{r}_{1}}^{2}=300\Rightarrow {{r}_{1}}=10\sqrt{3}\,\,\,\,\,\,\,\,\,(\text{Conventionally, }{{r}_{1}}>0)\\ &\Rightarrow \cos \theta =\dfrac{15}{10\sqrt{3}}=\dfrac{\sqrt{3}}{2}\text{ and }\sin \theta =\dfrac{5\sqrt{3}}{10\sqrt{3}}=\dfrac{1}{2}\\ &\Rightarrow \text{ Argument of }{{z}_{1}} = \theta =\dfrac{\pi }{6}\\ &{{z}_{2}}=-12{{i}^{7}}-4\sqrt{3}{{i}^{10}}=-12{{\left( {{i}^{2}} \right)}^{3}}i-4\sqrt{3}{{\left( {{i}^{2}} \right)}^{5}}=12i+4\sqrt{3} \\ & \Rightarrow {{r}_{2}}\cos \phi =4\sqrt{3},{{r}_{2}}\sin \phi =12\\ & \text{On squaring both sides of the equations and adding: }\\ & \Rightarrow {{r}_{2}}^{2}=192\Rightarrow {{r}_{2}}=8\sqrt{3}\,\,\,\,\,\,\,\,\,\,\,(\text{Conventionally, }{{r}_{2}}>0)\\ &\Rightarrow \cos \phi =\dfrac{4\sqrt{3}}{8\sqrt{3}}=\dfrac{1}{2}\text{ and }\sin \phi =\dfrac{12}{8\sqrt{3}}=\dfrac{\sqrt{3}}{2}\\ &\Rightarrow \text{Argument of }{{z}_{2}} = \phi =\dfrac{\pi }{3}\\ &\therefore \text{Required Argument = }\theta +\phi =\dfrac{\pi }{6}+\dfrac{\pi }{3}=\dfrac{\pi }{2} \end{align}$

 Convert the given complex number to polar form: $\dfrac{1-i}{\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}}$.

 Convert the given complex number in $a+ib$ form by rationalizing and then find modulus using formula $\sqrt{{{a}^{2}}+{{b}^{2}}}$. Find $\theta $ and then write in polar form as $z=r\left( \cos \theta +i\sin \theta  \right)$.

 $\begin{align} & \text{We have, }{{z}_{1}}=5\sqrt{3}i-15{{i}^{2}}=15+5\sqrt{3}i\,\,\,\,\,\,\,\,(\because {{i}^{2}}=-1)\\ & \text{Let }{{r}_{1}}\cos \theta =15\text{ and }{{r}_{1}}\sin \theta =5\sqrt{3}.\\ & \text{On squaring both sides of the equations and adding:}\\ & \Rightarrow {{r}_{1}}^{2}=300\Rightarrow {{r}_{1}}=10\sqrt{3}\,\,\,\,\,\,\,\,\,(\text{Conventionally, }{{r}_{1}} > 0)\\ & \Rightarrow \cos \theta =\dfrac{15}{10\sqrt{3}}=\dfrac{\sqrt{3}}{2}\text{ and }\sin \theta \\ & \Rightarrow \dfrac{5\sqrt{3}}{10\sqrt{3}}=\dfrac{1}{2}\\ & \Rightarrow \text{Argument of }{{z}_{1}}=\theta =\dfrac{\pi }{6}\\ &{{z}_{2}}=-12{{i}^{7}}-4\sqrt{3}{{i}^{10}}=-12{{\left( {{i}^{2}} \right)}^{3}}i-4\sqrt{3}{{\left( {{i}^{2}} \right)}^{5}}=12i+4\sqrt{3} \\ & \Rightarrow {{r}_{2}}\cos \phi =4\sqrt{3},{{r}_{2}}\sin \phi =12\\ & \text{On squaring both sides of the equations and adding:}\\ & \Rightarrow {{r}_{2}}^{2}=192\Rightarrow {{r}_{2}}=8\sqrt{3}\,\,\,\,\,\,\,\,\,(\text{Conventionally, }{{r}_{2}} > 0)\\ & \Rightarrow \cos \phi =\dfrac{4\sqrt{3}}{8\sqrt{3}}=\dfrac{1}{2}\text{ and }\sin \phi =\dfrac{12}{8\sqrt{3}}=\dfrac{\sqrt{3}}{2}\\ & \Rightarrow \text{Argument of }{{z}_{2}} = \phi =\dfrac{\pi }{3}\\ & \therefore \text{Required Argument = }\theta +\phi =\dfrac{\pi }{6}+\dfrac{\pi }{3}=\dfrac{\pi }{2} \end{align}$