Question

How to find the cube roots of `sqrt(2) + isqrt(2)` ? Exhibit them as vertices of certain regular polygon, and identify the principal root

Answer 1

`z^3=sqrt2+isqrt2`: `z=2e^(ipi/12),2e^(i(9pi)/12),2e^(i(17pi)/12)`. Principal root is `2e^(ipi/12)` and roots are vertices of triangle.

Explanation:

To solve `z^n=re^(itheta)`, we have `z=re^(i(theta/n+(2kpi)/n)), k={0,1,2,...,n-1}`.
Write `sqrt2+isqrt2` in the form `2e^(ipi/4)`. To find the principal root, raise to the power of `1/3` to get `2e^(ipi/12)`. To find the other two, add multiples of `2pi/3` to the exponent: `2e^(i(pi/12+(2pi)/3))` and `2e^(i(pi/12+(4pi)/3))`. In general, the nth roots of a complex number are the roots of a regular n-gon.

Answer 2

Please see the solutions below

Explanation:

The complex number is

`z=a+ib`

Here,

`z=sqrt2+isqrt2`

Transform this to the polar form

`z=|z|(costheta+isintheta)`

`{(|z|=sqrt(a^2+b^2)),(costheta=a/(|z|)),(sintheta=b/(|b|)):}`

Here,

`|z|=sqrt((sqrt2)^2+(sqrt2)^2)=sqrt4=2`

Therefore,

`z=2(sqrt2/2+isqrt2/2)`

`z=2(cos(pi/4+2kpi)+isin(pi/4+2kpi))`, `k in ZZ`

The cube root is

`z^(1/3)=(2(cos(pi/4+2kpi)+isin(pi/4+2kpi)))^(1/3)`

`=2^(1/3)(cos(1/3(pi/4+2kpi))+isin((1/3(pi/4+2kpi)))`

`=2^(1/3)(cos(pi/12+2/3kpi)+isin(pi/12+2/3kpi))`

When

`k=0`, `=>`, `z_1=2^(1/3)(cos(pi/12)+isin(pi/12))`

`k=1`, `=>`, `z_1=2^(1/3)(cos(3/4pi)+isin(3/4pi))`

`k=2`, `=>`, `z_1=2^(1/3)(cos(17/12pi)+isin(17/12pi))`