Question

find all the value of (1/2-i√3/2)^3/4 and hence prove that the product of the values is 1​

Answer 1

$\textbf{Given:}$

$\mathsf{\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)^\frac{3}{4}}$

$\textbf{To find:}$

$\textsf{All the values of}\;\mathsf{\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)^\frac{3}{4}}$

$\textbf{Solution:}$

$\textsf{First we convert the given complex number into polar form}$

$\mathsf{\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}=r[cos\theta+i\,sin\theta]}$

$\mathsf{Here,\;x=\dfrac{1}{2}\;\&\;y=\dfrac{-\sqrt{3}}{2}}}$

$\mathsf{r=\sqrt{x^2+y^2}=\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}=1}$

$\mathsf{cos\theta=\dfrac{x}{r}=\dfrac{1}{2}}$

$\mathsf{sin\theta=\dfrac{y}{r}=\dfrac{-\sqrt{3}}{2}}$

$\therefore\mathsf{\theta\;lies\;IV\;quadrant}$

$\implies\mathsf{\theta=\dfrac{-\pi}{3}}$

$\implies\mathsf{\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}=cos\left(\dfrac{-\pi}{3}\right)+i\,sin\left(\dfrac{-\pi}{3}\right)]}$

$\implies\mathsf{\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}=cos\left(2k\pi-\dfrac{\pi}{3}\right)+i\,sin\left(2k\pi-\dfrac{\pi}{3}\right)]}$

$\implies\mathsf{\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}=cos\left(\dfrac{(6k-1)\pi}{3}\right)+i\,sin\left(\dfrac{(6k-1)\pi}{3}\right)]}$

$\implies\mathsf{\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)\dfrac{3}{4}=[cos\left(\dfrac{(6k-1)\pi}{3}\right)+i\,sin\left(\dfrac{(6k-1)\pi}{3}\right)]^\frac{3}{4}}$

$\textsf{Apply demovire's theorem}$

$\implies\mathsf{\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)\dfrac{3}{4}=cos\left(\dfrac{3(6k-1)\pi}{4{\times}3}\right)+i\,sin\left(\dfrac{3(6k-1)\pi}{3{\times}4\right)}}$

$\implies\mathsf{\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)\dfrac{3}{4}=cos\left(\dfrac{(6k-1)\pi}{4}\right)+i\,sin\left(\dfrac{(6k-1)\pi}{4}\right)}\;\;i=0,1,2,3$

$\textsf{The values are}$

$\mathsf{k=0,\;\;cos\left(\dfrac{\pi}{4}\right)+i\,sin\left(\dfrac{-\pi}{4}\right)=e^\frac{-i\pi}{4}}$

$\mathsf{k=1,\;\;cos\left(\dfrac{5\pi}{4}\right)+i\,sin\left(\dfrac{5\pi}{4}\right)=e^\frac{i5\pi}{4}}$

$\mathsf{k=2,\;\;cos\left(\dfrac{11\pi}{4}\right)+i\,sin\left(\dfrac{11\pi}{4}\right)=e^\frac{i11\pi}{4}}$

$\mathsf{k=3,\;\;cos\left(\dfrac{17\pi}{4}\right)+i\,sin\left(\dfrac{17\pi}{4}\right)=e^\frac{i17\pi}{4}}$

$\implies\mathsf{Product\;of\;the\;values}$

$\mathsf{=e^\frac{-i\pi}{4}{\times}e^\frac{i5\pi}{4}{\times}e^\frac{i11\pi}{4}{\times}e^\frac{i17\pi}{4}}$

$\mathsf{=e^\frac{i32\pi}{4}}$

$\mathsf{=e^{i8\pi}}$

$\mathsf{=(e^{i\,\pi})^8}$

$\mathsf{=(cos\pi+i\,sin\pi)^8}$

$\mathsf{=(-1+i\,0)^8}$

$\mathsf{=(-1)^8}$

$\mathsf{=1}$

$\textbf{Find more:}}$

Solve the following equations in complex numbers and write your answer in polar and rectangular form

z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z = 0

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