Question

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

Answer 1

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 :)

Answer 2

`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`

Answer 3

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)`