Question

In Problems 53–60, find all the complex roots. Leave your answers in polar form with the argument in degrees.

54. The complex fourth roots of$\sqrt{3}-i$.

Short answer

The complex four roots of$\sqrt{3}-i$are$\sqrt[4]{2}(\cos 82.5+i\sin 82.5),\sqrt[4]{2}(\cos 172.5+i\sin 172.5),\sqrt[4]{2}(\cos 262.5+i\sin 262.5),\sqrt[4]{2}(\cos 352.5+i\sin 352.5)$.

Step-by-step solution

Step 1. Rewrite z=3-i in the polar form.

The magnitude of $\sqrt{3}-i$is

$$\begin{gathered} \sqrt{{\left(\sqrt{3}\right)}^2+{(-1)}^2}=\sqrt{3+1} \\ =\sqrt{4} \\ =2 \end{gathered}$$

which gives

$$\begin{gathered} \sqrt{3}-i=2(\frac{\sqrt{3}}{2}-i\frac{1}{2}) \\ =2(\cos 330°+i\sin 330°) \end{gathered}$$

Step 2. Using De Moivre's Theorem.

We get

$$\begin{gathered} {(\sqrt{3}-i)}^{\frac{1}{4}}={\left[2\right(\cos 330°+i\sin 330°\left)\right]}^{\frac{1}{4}} \\ {z}_k=\left\{\right(2{)}^{\frac{1}{4}}\left[\cos \right(\frac{330}{4}+\frac{360k}{4})+i\sin (\frac{330}{4}+\frac{360k}{4}\left)\right]\};k=0,1,2,3 \\ =\sqrt[4]{2}[\cos (82.5+90k)+i\sin (82.5+90k)];k=0,1,2,3 \end{gathered}$$

Step 3. Substitute k=0,1,2,3 in zk=24[cos(82.5+90k)+isin(82.5+90k)].

We get

$$\begin{gathered} z_0=\sqrt[4]{2}(\cos 82.5+i\sin 82.5) \\ {z}_1=\sqrt[4]{2}\left(\cos \right(82.5+90)+\sin (82.5+90\left)\right) \\ =\sqrt[4]{2}(\cos 172.5+i\sin 172.5) \\ {z}_2=\sqrt[4]{2}\left[\cos \right(82.5+180)+i\sin (82.5+180\left)\right] \\ =\sqrt[4]{2}(\cos 262.5+i\sin 262.5) \\ {z}_3=\sqrt[4]{2}\left[\cos \right(82.5+270)+i\sin (82.5+270\left)\right] \\ =\sqrt[4]{2}(\cos 352.5+i\sin 352.5) \end{gathered}$$

So we have found the complex fourth roots of$\sqrt{3}-i$.