Question

1) Convert 2-7i to trigonometric form

2) Use the n-th roots theorem to find the requested roots of the given complex number.
Find the cube roots of 125

Answer 1

1)
`√(53)(\cos286^\circ+i\sin286^\circ)`

2)
`\displaystyle 5,-(5)/(2)+(5√(3))/(2)i,-(5)/(2)-(5√(3))/(2)i`

Explanation:

Problem 1

`z=2-7i\\\\r=√(a^2+b^2)=√(2^2+(-7)^2)=√(4+49)=√(53)\\\\\theta=\tan^(-1)((y)/(x))=\tan^(-1)((-7)/(2))\approx-74^\circ=360^\circ-74^\circ=286^\circ\\\\z=r\,(\cos\theta+i\sin\theta)=√(53)(\cos286^\circ+i\sin 286^\circ)`

Problem 2

`\displaystyle z^(1)/(n)=r^(1)/(n)\biggr[\text{cis}\biggr((\theta+2k\pi)/(n)\biggr)\biggr]\,\,\,\,\,\,\,k=0,1,2,3,\,...\,,n-1\\\\z^(1)/(3)=125^(1)/(3)\biggr[\text{cis}\biggr((0+2(2)\pi)/(3)\biggr)\biggr]=5\,\text{cis}\biggr((4\pi)/(3)\biggr)=5\biggr(-(1)/(2)-(√(3))/(2)i\biggr)=-(5)/(2)-(5√(3))/(2)i`

`\displaystyle z^(1)/(3)=125^(1)/(3)\biggr[\text{cis}\biggr((0+2(1)\pi)/(3)\biggr)\biggr]=5\,\text{cis}\biggr((2\pi)/(3)\biggr)=5\biggr(-(1)/(2)+(√(3))/(2)i\biggr)=-(5)/(2)+(5√(3))/(2)i`

`\displaystyle z^(1)/(3)=125^(1)/(3)\biggr[\text{cis}\biggr((0+2(0)\pi)/(3)\biggr)\biggr]=5\,\text{cis}(0)=5(1+0i)=5`

Note that
`\text{cis}\,\theta=\cos\theta+i\sin\theta` and
`125=125(\cos0^\circ+i\sin0^\circ)`