Find the cube root of complex number ? Z=-√2 + i√2
3 Answers
multiply and divide LHS by -2
we get
therefore
hence cube root of
hope u find it helpful :)
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^(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#
The solutions are
Explanation:
Re write
Therefore,
When
When
When