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

2 Answers
Teddy
Jul 4, 2018

#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.

Narad T.
Jul 4, 2018

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))#

Related questions
Impact of this question
3360 views around the world
You can reuse this answer
Creative Commons License