# z1= 1+ √3 * i Apply De Moivere's theorem to find the modulus and angles of the roots of z1. Please help.

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### 1 Answer

Given `z=1+isqrt(3)` we are asked to describe the angles and moduli of the roots:

(1) First we write in polar form: if `z=a+bi` then `r=sqrt(a^2+b^2)` and then `z=rcostheta+(rsintheta)i` or `rcistheta` .

We can find `theta` using trig functions: `tantheta=sqrt(3)/1==>theta=pi/3`

`r=sqrt((1)^2+sqrt(3)^2)=2`

So `z=2cispi/3` or `z=2cis60^@`

(2) DeMoivre's theorem is if `z=rcistheta` then `z^n=r^ncisntheta`

(3) You do not specify the roots. In the general case the nth root of z is given by `root(n)(z)=z^(1/n)=r^(1/n)cis(theta/n+(k*2pi)/n),k in ZZ^+,0<=k<=n-1`

For example: **the square roots of z are:**

`2^(1/2)cis(pi/6+kpi)=sqrt(2)cispi/6,sqrt(2)cis(7pi)/6` The modulus is `sqrt(2)` and the angles are `pi/6,(7pi)/6`

**The cube roots of z:**

`2^(1/3)cis(pi/9+(2kpi)/3)=root(3)(2)cispi/9,root(3)(2)cis(4pi)/9,root(3)(2)cis(7pi)/9` The modulus is `root(3)(2)` and the angles are `pi/9,(4pi)/9,(7pi)/9` etc...