Question

I need help with this practice problem solving In addition, after you review what branch of mathematics do you think this problem is? Just curious

Answer 1

Given

The equation,

$x^3=-4+4i$

To find all the solutions of the given equation in polar form.

Explanation:

It is given that,

$x^3=-4+4i$

That implies,

$\begin{gathered} x=(-4+4i)^{\frac{1}{3}} \\ x=(4)^{\frac{1}{3}}(-1+i)^{\frac{1}{3}} \end{gathered}$

Now, consider

$\begin{gathered} z=-1+i \\ \Rightarrow r=\sqrt{x^2+y^2} \\ =\sqrt{(-1)^2+1^2} \\ =\sqrt{1+1} \\ =\sqrt{2} \end{gathered}$

Also,

$\begin{gathered} \theta=\tan^{-1}|\frac{y}{x}| \\ =\tan^{-1}|\frac{1}{-1}| \\ =\tan^{-1}|-1| \\ =\tan^{-1}(1) \\ =\frac{\pi}{4} \end{gathered}$

Since (x,y) lies in 2nd quadrant.

Then,

$\begin{gathered} \varphi=\pi-\theta \\ =\pi-\frac{\pi}{4} \\ =\frac{3\pi}{4} \end{gathered}$

Therefore,

$\begin{gathered} z=r(\cos\varphi+i\sin\varphi) \\ =\sqrt{2}e^{i(\frac{3\pi}{4})} \end{gathered}$

That implies,

$\begin{gathered} x=(4)^{\frac{1}{3}}\lbrace(\sqrt{2}e^^{i(\frac{3\pi}{4})})\rbrace^{\frac{1}{3}} \\ =(2^2)^{\frac{1}{3}}(2^{\frac{1}{2}})^{\frac{1}{3}}(e^{i(\frac{3\pi}{4})})^{\frac{1}{3}} \\ =\sqrt[6]{32}(e^{i(\frac{3\pi}{4})})^{\frac{1}{3}} \\ =\sqrt[6]{32}\lbrace cis(\frac{3\pi}{4})\rbrace^{\frac{1}{3}} \end{gathered}$

Now,

$Add\text{ }2k\pi\text{ to }\varphi=\frac{3\pi}{4}$

That implies,

$x=\sqrt[6]{32}\lbrace cis(\frac{3\pi}{4}+2k\pi)\rbrace^{\frac{1}{3}}$

By applying De-movers theorem,

$x=\sqrt[6]{32}\lbrace cis\frac{1}{3}(\frac{3\pi}{4}+2k\pi)\rbrace$

Put k=0,1,2 in the above equation.

That implies,

$\begin{gathered} When\text{ }k=0, \\ x=\sqrt[6]{32}\lbrace cis\frac{1}{3}(\frac{3\pi}{4})\rbrace \\ =\sqrt[6]{32}\lbrace cis(\frac{\pi}{4})\rbrace \\ When\text{ }k=1, \\ x=\sqrt[6]{32}\lbrace cis\frac{1}{3}(\frac{3\pi}{4}+2\pi)\rbrace \\ =\sqrt[6]{32}\lbrace cis\frac{1}{3}(\frac{3\pi+8\pi}{4})\rbrace \\ =\sqrt[6]{32}\lbrace cis(\frac{11\pi}{12})\rbrace \\ When\text{ }k=2, \\ x=\sqrt[6]{32}\lbrace cis\frac{1}{3}(\frac{3\pi}{4}+4\pi)\rbrace \\ =\sqrt[6]{32}\lbrace cis\frac{1}{3}(\frac{3\pi+16\pi}{4})\rbrace \\ =\sqrt[6]{32}\lbrace cis(\frac{19\pi}{12})\rbrace \end{gathered}$

Hence, the solutions are,

$x=\sqrt[6]{32}\lbrace cis(\frac{\pi}{4})\rbrace,\sqrt[6]{32}\lbrace cis(\frac{11\pi}{12})\rbrace,\sqrt[6]{32}\lbrace cis(\frac{19\pi}{12})\rbrace$