Question

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

53. The complex cube root of$1+i$.

Short answer

The complex cube roots of$1+i$arerole="math" localid="1646812515277" $\sqrt[6]{2}(\cos 15+i\sin 15),\sqrt[6]{2}(\cos 135+i\sin 135),\sqrt[6]{2}(\cos 255+i\sin 255)$.

Step-by-step solution

Step 1. Rewrite 1+i in the polar form.

The magnitude of $1+i$is

$\sqrt{1^2+1^2}=\sqrt{2}$

which gives

$$\begin{gathered} 1+i=\sqrt{2}(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}) \\ =\sqrt{2}(\cos 45°+i\sin 45°) \end{gathered}$$

Step 2. Using the theorem of complex roots.

We get

$$\begin{gathered} z_k=\sqrt[3]{\sqrt{2}}\left[\cos \right(\frac{45}{3}+\frac{360k}{3})+i\sin (\frac{45}{3}+\frac{360k}{3}\left)\right];k=0,1,2 \\ =\sqrt[6]{2}\left[\cos \right(15+120k)+i\sin (15+120k\left)\right] \end{gathered}$$

Step 3. Substitute k=0,1,2 in zk=26(cos(15°+120°k)+isin(15°+120°k))

We get

$$\begin{gathered} z_0=\sqrt[6]{2}(\cos 15+i\sin 15) \\ {z}_1=\sqrt[6]{2}\left(\cos \right(15+120)+i\sin (15+120\left)\right) \\ =\sqrt[6]{2}(\cos 135+i\sin 135) \\ {z}_2=\sqrt[6]{2}\left(\cos \right(15+240)+i\sin (15+240\left)\right) \\ =\sqrt[6]{2}(\cos 255+i\sin 255) \end{gathered}$$

So we have found the complex cube roots of$1+i$.