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.

Step-by-step explanation:

It is given that,

`x^3=-4+4i`

That implies,

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

Now, consider

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

Also,

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

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

Then,

`\begin{gathered} \varphi=\pi-\theta \\ =\pi-(\pi)/(4) \\ =(3\pi)/(4) \end{gathered}`

Therefore,

`\begin{gathered} z=r(\cos\varphi+i\sin\varphi) \\ =√(2)e^{i((3\pi)/(4))} \end{gathered}`

That implies,

Now,

`Add\text{ }2k\pi\text{ to }\varphi=(3\pi)/(4)`

That implies,

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

By applying De-movers theorem,

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

Hence, the solutions are,

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