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Graphing Complex Numbers Quiz 2

mathematics

 Convert the given complex number to polar form: $-2$.

 If $z$ is a complex number then the polar form of $z$ is represented by: $z=r\left( \cos \theta +i\sin \theta  \right)$.

  $\begin{align} & \text{Given complex number: }-2+0i\\ & \text{Let }r\cos \theta =-2\text{ and }r\sin \theta =0\\ & \text{On squaring on both sides and adding the equations, we obtain} \\ & \Rightarrow {{r}^{2}}=4\Rightarrow r=2\\ &\\ &(\text{Conventionally}, r > 0)\\ &\therefore 2\cos \theta =-2\text{ and }2\sin \theta =0\\ &\Rightarrow \cos \theta =-1\text{ and }\sin \theta =0\\ &\Rightarrow \theta =\pi\\ & \text{So, polar form of }-2\text{ is: }2\left( \cos \pi +i\sin \pi  \right)\end{align}$

 Convert the given complex number to polar form: $-i$

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

 Find the modulus and the argument of the complex number $z=1-\sqrt{3}i$.

 If $z$ is a complex number then the polar form of $z$ is represented by: $z=r\left( \cos \theta +i\sin \theta  \right)$. $r$ represents modulus and $\theta $ represents Argument. 

 $\begin{align} & \text{We have, }z=1-\sqrt{3}i\\ & \text{Let }r\cos \theta =1\text{ and }r\sin \theta =-\sqrt{3}.\\ & \text{On squaring both sides of the equations and adding: }\\ & \Rightarrow {{r}^{2}}=4\Rightarrow r=2\\ &\\ &(\text{Conventionally}, r>0)\\ & \therefore \text{Modulus} = 2\\ &\Rightarrow \cos \theta =\dfrac{1}{2}\text{ and }\sin \theta =\dfrac{-\sqrt{3}}{2}\\ & \text{Since the value of } “\cos \theta” \text{ is positive and } “\sin \theta” \text{ is negative then they are in IV quadrant,}\\ &\therefore \text{Argument }= \dfrac{-\pi }{3}\end{align}$

 Represent the complex number $z=-1+\sqrt{3}i$ in the polar form.

 $\begin{align} & \text{We have: }z=-1+\sqrt{3}i\\ & \text{Let }r\cos \theta =-1\text{ and }r\sin \theta =\sqrt{3}\\ & \text{On squaring on both sides and adding the equations, we obtain }\\ & \Rightarrow {{r}^{2}}=4\Rightarrow r=2\\ &\\ &(\text{Conventionally}, r > 0)\\ & \Rightarrow \cos \theta =\dfrac{-1}{2}\text{ and }\sin \theta =\dfrac{\sqrt{3}}{2}\\ &\Rightarrow \theta =\left( \pi -\dfrac{\pi }{3} \right)=\dfrac{2\pi }{3}\\ & \text{Therefore, required polar form: }z=2\left( \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} \right)\end{align}$

 Represent the complex number $z=\sin \dfrac{2\pi }{3}-i\cos \dfrac{2\pi }{3}$in the polar form.

 $\begin{align} & \text{We have: }z=\sin \dfrac{2\pi }{3}-i\cos \dfrac{2\pi }{3}\\ &\Rightarrow z=\dfrac{\sqrt{3}}{2}-i\left( \dfrac{-1}{2} \right)\\ &\Rightarrow \dfrac{\sqrt{3}}{2}+i\left( \dfrac{1}{2} \right)\\ &\text{Let }r\cos \theta =\dfrac{\sqrt{3}}{2}\text{ and }r\sin \theta =\dfrac{1}{2}\\ & \text{On squaring on both sides and adding the equations, we obtain }\\ &\Rightarrow r=1\\ &\\ &(\text{Conventionally}, r > 0)\\ &\Rightarrow \cos \theta =\dfrac{\sqrt{3}}{2}\text{ and }\sin \theta =\dfrac{1}{2}\\ &\Rightarrow \theta =\dfrac{\pi }{6}\\ &\text{Therefore, required polar form: }z=1\left( \cos \dfrac{\pi }{6}+i\sin \dfrac{\pi }{6} \right)\end{align}$