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Polar system Quiz 2

mathematics

 Show $\sqrt{5}i$ using polar coordinates.

 First find the value of $r$ by the formula $r=\sqrt{{{a}^{2}}+{{b}^{2}}}$ for the complex number $a+ib$ then find $\theta $ by using $x=r\cos \theta $.

  On the complex plane, the number $z=\sqrt{5}i$ is the same as $z=0+\sqrt{5}i$ $\\ $ Writing it in polar form, we have to calculate $r$ first. $ \\ r=\sqrt{{{0}^{2}}+{{\sqrt{5}}^{2}}}=\sqrt{5} \\ $ If $x=r\cos \theta$ and $x=0$, then $\theta =\dfrac{\pi }{2}$ $ \\ $ In polar coordinates, the complex number $z=0+\sqrt{5}i$ can be written as, $ \\ z=\sqrt{5}\left( \cos \left( \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\pi }{2} \right) \right)$

 Find the rectangular form of the equation whose polar form is : $z=3\left( cos\left( \dfrac{2\pi }{3} \right)+isin\left( \dfrac{2\pi }{3} \right) \right)$

 First substitute the trigonometric values in the expression then apply distributive property.

  $ z=3\left( \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right) \\ $ By evaluating the trigonometric expressions, we get, $ \\ \text{cos}\left( \dfrac{2\pi }{3} \right)=-\cos \left( \dfrac{\pi }{3} \right)=-\dfrac{1}{2}\text{ and sin}\left( \dfrac{2\pi }{3} \right)=\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\ $ After substitution, the complex number becomes $ \\ \Rightarrow z=3\left( -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right) \\ $ Applying the distributive property: $ \\ z=-\dfrac{3}{2}+\dfrac{3\sqrt{3}}{2}i$

 Find the product of the two complex numbers, $11\left( \cos 212{}^\circ +i\sin 212{}^\circ  \right)$ and  $13\left( \cos 13{}^\circ +i\sin 13{}^\circ  \right)$.

 Use the multiplication property ${{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left[ A+iB\right]$.

${{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left[ A+iB\right]\right] \\  \Rightarrow {{z}_{1}}{{z}_{2}}=11\times 13\left[ \cos 225{}^\circ +i\sin 225{}^\circ  \right] \\  \Rightarrow {{z}_{1}}{{z}_{2}}=143\left[ \cos \left( \dfrac{5\pi }{4} \right)+i\sin \left( \dfrac{5\pi }{4} \right) \right] \\  \Rightarrow {{z}_{1}}{{z}_{2}}=143\left[ -\dfrac{\sqrt{2}}{2}+i\left( -\dfrac{\sqrt{2}}{2} \right) \right] \\  \Rightarrow {{z}_{1}}{{z}_{2}}=-143\left( \dfrac{\sqrt{2}}{2} \right)\left( 1+i \right)$

 Evaluate: ${{\left( \sqrt{3}+\sqrt{3}i \right)}^{7}}$

 First convert the complex number into polar form then by using De Moivre’s Theorem ${{(a+ib)}^{n}}={{r}^{n}}\left[ \cos (n\theta )+i\sin (n\theta ) \right]$, find the required value.

 Let's solve by using De Moivre's Theorem, which applies to complex numbers written in polar form, we must first write $ \\ $ $\left( \sqrt{3}+\sqrt{3}i \right)$ in polar form $ \\  r=\sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}=\sqrt{6} \\ $ Then we find $\theta$ $ \\ $ Using the formula $\tan \theta =\dfrac{y}{x}$ gives $\tan \theta =\dfrac{1}{1}=1$ $ \\  \Rightarrow \theta =\dfrac{\pi }{4} \\ $ Using De Moivre's Theorem to evaluate the expression, $ \\ {{(a+ib)}^{n}}={{r}^{n}}\left[ \cos (n\theta )+i\sin (n\theta ) \right] \\  \Rightarrow {{\left( \sqrt{3}+\sqrt{3}i \right)}^{7}}={{\left( \sqrt{6} \right)}^{7}}\left[ \cos \left( 5\times \dfrac{\pi }{4} \right)+i\sin \left( 5\times \dfrac{\pi }{4} \right) \right] \\  \Rightarrow {{\left( \sqrt{3}+\sqrt{3}i \right)}^{7}}={{\left( \sqrt{6} \right)}^{7}}\left[ -\dfrac{\sqrt{2}}{2}+i\left( -\dfrac{\sqrt{2}}{2} \right) \right] \\  \Rightarrow {{\left( \sqrt{3}+\sqrt{3}i \right)}^{7}}={{\left( \sqrt{6} \right)}^{7}}\left( \dfrac{\sqrt{2}}{2} \right)(1+i)$

 Find a possible value of $\sqrt[3]{z}$ among the following, if $z=\dfrac{{{z}_{1}}}{{{z}_{2}}}$ such that ${{z}_{1}}=54\left( \cos 169{}^\circ +i\sin 169{}^\circ  \right)\text{ and }{{z}_{2}}=2\left( \cos 49{}^\circ +i\sin 49{}^\circ  \right)$

 First use the division property $\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}\left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right],{{z}_{2}}\ne 0$then use ${{n}^{th}}$ root Theorem,${{z}^{\dfrac{1}{n}}}={{r}^{\dfrac{1}{n}}}\left[ \cos \left( \dfrac{\theta }{n}+\dfrac{2k\pi }{n} \right)+i\sin \left( \dfrac{\theta }{n}+\dfrac{2k\pi }{n} \right) \right],k=0,1,2,3,...,n-1$.

 $z=\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{r}_{1}}}{{{r}_{2}}}\left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right],{{z}_{2}}\ne 0 \\  z=27\left[ \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right] \\  z=27\left[ \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right] \\  {{z}^{\dfrac{1}{3}}}={{27}^{\dfrac{1}{3}}}\left[ \cos \left( \dfrac{\dfrac{2\pi }{3}}{3}+\dfrac{2k\pi }{3} \right)+i\sin \left( \dfrac{\dfrac{2\pi }{3}}{3}+\dfrac{2k\pi }{3} \right) \right]=3\left[ \cos \left( \dfrac{2\pi }{9}+\dfrac{2k\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{9}+\dfrac{2k\pi }{3} \right) \right] \\  \text{There will be three roots: }k=0,1,2. \\  \text{When }k=0,{{z}^{\dfrac{1}{3}}}=3\left[ \cos \left( \dfrac{2\pi }{9} \right)+i\sin \left( \dfrac{2\pi }{9} \right) \right] \\  \text{When }k=2,{{z}^{\dfrac{1}{3}}}=3\left[ \cos \left( \dfrac{2\pi }{9}+\dfrac{12\pi }{9} \right)+i\sin \left( \dfrac{2\pi }{9}+\dfrac{12\pi }{9} \right) \right] \\  \Rightarrow {{z}^{\dfrac{1}{3}}}=3\left[ \cos \left( \dfrac{14\pi }{9} \right)+i\sin \left( \dfrac{14\pi }{9} \right) \right] \\  \text{When }k=1,{{z}^{\dfrac{1}{3}}}=3\left[ \cos \left( \dfrac{2\pi }{9}+\dfrac{6\pi }{9} \right)+i\sin \left( \dfrac{2\pi }{9}+\dfrac{6\pi }{9} \right) \right]=3\left[ \cos \left( \dfrac{8\pi }{9} \right)+i\sin \left( \dfrac{8\pi }{9} \right) \right] \\ $