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Complex Numbers, EXT2 N1 2022 HSC 12e

Given the complex number  `z=e^(i theta)`, show that  `w=(z^(2)-1)/(z^(2)+1)`  is purely imaginary.  (3 marks)

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`text{Proof (See Worked Solutions)}`

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`z=e^(i theta)`

`w` `=(z^2-1)/(z^2+1)`  
  `=(e^(i2theta)-1)/(e^(i2theta)+1)`  
  `=(e^(i2theta)-1)/(e^(i2theta)+1) xx (e^(-i2theta)+1)/(e^(-i2theta)+1)`   
  `=(1+e^(i2theta)-e^(-i2theta)-1)/(1+e^(i2theta)+e^(-i2theta)+1)`  
  `=(e^(i2theta)-e^(-i2theta))/(2+e^(i2theta)+e^(-i2theta))`  
  `=(cos(2theta)+isin(2theta)-(cos(-2theta)+isin(-2theta)))/(2+cos(2theta)+isin(2theta)+cos(-2theta)+isin(-2theta))`  
  `=(cos(2theta)+isin(2theta)-cos(2theta)+isin(2theta))/(2+cos(2theta)+isin(2theta)+cos(2theta)-isin(2theta))`  
  `=(i2sin(2theta))/(2+2cos(2theta))`  
  `=i((2sin(2theta))/(2+2cos(2theta)))\ \ text{(purely imaginary)}`  

Complex Numbers, EXT2 N1 2022 HSC 11c

  1. Write the complex number  \(-\sqrt{3}+i\) in exponential form.   (2 marks)

  2. Hence, find the exact value of  \((-\sqrt{3}+i)^{10}\) giving your answer in the form  \(x+i y\).   (2 marks)

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i.    \(2 e^{\small{\dfrac{5 \pi}{6}} i}\)

ii.   \(512+512 \sqrt{3} i\)

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i.   

\(\text {Let}\ \ z=-\sqrt{3}+i\)

\(\abs{z}=\sqrt{(-\sqrt{3})^2+1^2}=2\)

\(\text{Find}\ \ \arg (z):\)

\(\tan \theta=\dfrac{1}{\sqrt{3}} \Rightarrow \theta=\dfrac{\pi}{6}\)

\(\Rightarrow \arg (z)=\dfrac{5 \pi}{6}\)

\(\therefore z=2\left(\dfrac{\cos (5 \pi)}{6}+\dfrac{\sin (5 \pi)}{6} i\right)=2 e^{\small{\dfrac{5 \pi}{6}} i}\)
 

ii.    \((-\sqrt{3}+i)^{10}\) \(=\left(2 e^{\small{\dfrac{5 \pi}{6}} i}\right)^{10}\)
    \(=2^{10} e^{\small{\dfrac{50 \pi}{6}} i}\)
    \(=1024 e^{\small{\dfrac{\pi}{3}} i}\)
    \(=1024\left(\cos \left(\dfrac{\pi}{3}\right)+\sin \left(\dfrac{\pi}{3}\right) i\right)\)
    \(=1024\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} i\right)\)
    \(=512+512 \sqrt{3} i\)

Complex Numbers, EXT2 N1 SM-Bank 1 MC

Which of the following is the complex number  \(-\sqrt{3}+3 i ?\)?

  1. \(2 \sqrt{3} e^{-\small{\dfrac{i \pi}{3}}}\)
  2. \(2 \sqrt{3} e^{\small{\dfrac{i 2 \pi}{3}}}\)
  3. \(12 e^{-\small{\dfrac{i \pi}{3}}}\)
  4. \(-12 e^{\small{\dfrac{i 2 \pi}{3}}}\)
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\(B\)

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\(\abs{z}\) \(=\sqrt{(\sqrt{3})^2+3^2}=2 \sqrt{3}\)
\(\tan \theta\) \(=\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}}\)
\(\theta\) \(=\dfrac{\pi}{6}\)

\(\arg (z)=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{2 \pi}{3}\)
  

\(\therefore z\) \(=2 \sqrt{3}\left(\cos \left(\dfrac{2 \pi}{3}\right)+i \sin \left(\dfrac{2 \pi}{3}\right)\right.\)
  \(=2 \sqrt{3} e^{\small{\dfrac{i 2 \pi}{3}}}\)

\(\Rightarrow B\)

Complex Numbers, EXT2 N1 EQ-Bank 3 MC

Which of the following is the complex number  \(-3-\sqrt{3}i\)?

  1. \(12 e^{-\small{\dfrac{i 5 \pi}{6}}}\)
  2. \(12 e^{\small{\dfrac{i \pi}{6}}}\)
  3. \(2 \sqrt{3} e^{\small{-\dfrac{i 5 \pi}{6}}}\)
  4. \(2 \sqrt{3} e^{\small{\dfrac{i \pi}{6}}}\)
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\(C\)

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\(\abs{z}\) \(=\sqrt{3^2+(\sqrt{3})^2}\)
  \(=\sqrt{12}\)
  \(=2 \sqrt{3}\)

 

\(\tan \theta\) \(=\dfrac{3}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}=\sqrt{3}\)
\(\theta\) \(=\dfrac{\pi}{3}\)
\(\arg (z)\) \(=-\left(\dfrac{\pi}{3}+\dfrac{\pi}{2}\right)=-\dfrac{5 \pi}{6}\)

\(\text {In exponential form:}\)

\(z=2 \sqrt{3} e^{-\small{\dfrac{i5\pi}{6}}}\)

\(\Rightarrow C\)

Complex Numbers, EXT2 N1 EQ-Bank 12

Express the complex number  \(z=-2 \sqrt{2}-2 \sqrt{6} i\)  in the exponential form.   (2 marks)

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\(4 \sqrt{2} e^{-i \small{\dfrac{2 \pi}{3}}}\)

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\(\abs{z}\) \(=\sqrt{(2 \sqrt{2})^2+(2 \sqrt{6})^2}\)
  \(=\sqrt{8+24}\)
  \(=4 \sqrt{2}\)

 

\(\tan \theta\) \(=\dfrac{2 \sqrt{2}}{2 \sqrt{6}}=\dfrac{1}{\sqrt{3}}\)
\(\theta\) \(=\dfrac{\pi}{6}\)

 

\(\operatorname{Arg}(z)=-\left(\dfrac{\pi}{2}+\dfrac{\pi}{6}\right)=-\dfrac{2 \pi}{3}\)

\(\therefore z\) \(=4 \sqrt{2} \operatorname{cis}\left(\dfrac{-2 \pi}{3}\right)\)
  \(=4 \sqrt{2} e^{-i \small{\dfrac{2 \pi}{3}}}\)