Find the cube root of complex number ? Z=-√2 + i√2

3 Answers
Jan 17, 2018

multiply and divide LHS by -2
we get
#z = -2( 1/2^(1/2) - i1/2^(1/2)) = -2*e^(i-pi/4) = |z|*e^(i-pi/4)#
therefore
#z^(1/3) = (|z|*e^(i-pi/4))^(1/3)#
#=-2^(1/3)*e^(i-pi/12)#
#=-2^(1/3)(cos(-pi/12) + isin(-pi/12))#
hence cube root of #z# is
#-2^(1/3)(cos(-pi/12) + isin(-pi/12))#
hope u find it helpful :)

Jan 17, 2018

#root(3)(z) = 2^(-1/6)+2^(-1/6)i#

Explanation:

Given:

#z = -sqrt(2)+i sqrt(2)#

#color(white)(z) = 2(cos((3pi)/4)+ i sin((3pi)/4))#

The principal cube root of #z# is:

#z^(1/3) = 2^(1/3)(cos(pi/4)+i sin(pi/4))#

#color(white)(z^(1/3)) = 2^(1/3)(sqrt(2)/2+i sqrt(2)/2)#

#color(white)(z^(1/3)) = 2^(-1/6)+2^(-1/6) i#

Jan 18, 2018

The solutions are #S={2^(1/6)(1+i), 2^(1/3)(-(sqrt2+sqrt6)/4+i(sqrt6-sqrt2)/4), 2^(1/3)(-(sqrt2+sqrt6)/4-i(sqrt6+sqrt2)/4)}#

Explanation:

Re write #z# in the exponential form

#z=-sqrt2+isqrt2=sqrt2(-1+i)#

#=2(-1/sqrt2+1/sqrt2i)#

#=-2(cos(3/4pi)+isin(3/4pi))#

#=2e^(3/4pi+2kpi)#, #k in ZZ #

Therefore,

#z^(1/3)=2^(1/3)e^(i(1/4pi+2/3kpi)#

When #k=0#

#z_0=2^(1/3)e^(i1/4pi)=2^(1/3)(cos(pi/4)+isin(pi/4))#

#=2^(1/6)(1+i)#

When #k=1#

#z_1=2^(1/3)e^(i11/12pi)=2^(1/3)(cos(11/12pi)+isin(11/12pi))#

#=2^(1/3)(-(sqrt2+sqrt6)/4+i(sqrt6-sqrt2)/4)#

When #k=2#

#z_2=2^(1/3)e^(i19/12pi)=2^(1/3)(cos(19/12pi)+isin(19/12pi))#

#=2^(1/3)(-(sqrt2+sqrt6)/4-i(sqrt6+sqrt2)/4)#