Hi, I'm stuck on this question and would like help. Thanks if anyone can explain.
View attachment 37310
Step 1 Draw an Argand diagram with the complex number specified
Step 2 Write in mod-arg form.
![](https://latex.codecogs.com/png.latex?\bg_white \sqrt{3}+i=2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\right))
Step 3 Rewrite step 2 but instead modify the mod-arg form so that it looks like this
![](https://latex.codecogs.com/png.latex?\bg_white \sqrt{3}+i=2\left(\cos\left(\frac{\pi}{6}+2n\pi\right)+i\sin\left(\frac{\pi}{6}+2n\pi\right)\right)=2\left(\cos\left(\frac{\pi+12n\pi}{6}\right)+isin\left(\frac{\pi+12n\pi}{6}\right)\right))
where n is an integer
Step 4 Since this is cube root we divide the angle by 3 which is a reversal of De Moivre's Theorem and cube root the 2 giving us
![](https://latex.codecogs.com/png.latex?\bg_white \sqrt[3]{2}\left(\cos\left(\frac{\pi+12n\pi}{18}\right)+isin\left(\frac{\pi+12n\pi}{18}\right)\right))
Step 5 All the cube roots have to be within the domain of
![](https://latex.codecogs.com/png.latex?\bg_white -\pi\leq{\theta}\leq{\pi})
so using what we know we will then have these three answers
![](https://latex.codecogs.com/png.latex?\bg_white \sqrt[3]{2}\left(\cos\left(\frac{\pi}{18}\right)+isin\left(\frac{\pi}{18}\right)\right),\;\sqrt[3]{2}\left(\cos\left(\frac{\13\pi}{18}\right)+isin\left(\frac{13\pi}{18}\right)\right),\;\sqrt[3]{2}\left(\cos\left(\frac{-11\pi}{18}\right)+isin\left(\frac{-11\pi}{18}\right)\right))
Step 6 Put them on the Argand diagram
Step 7 Use Pythagoras Theorem and trigonometry in conjunction with the area of the triangle to solve the problem.