Question

A complex valued function f(z) =z3−1−I, what is the solution of f (z) =0? Note: exp(z) = e^z

Answer 1

A complex valued function f(z) =z3−1−I, what is the solution of f (z) =0? Note: exp(z) = e^z

Assuming you mean “I” to be the imaginary number unit i, this problem is actually easy to visualize on the complex plane.

 

z^3 = 1 + i

 

describes finding the cube root of 1 + i . That (1 + i) is a position at a 45 degree angle up from the x-axis, and square_root(2) away from the origin, i.e. 2^(0.5)*e^(i.pi./4). To take the cube root, you take the cube root of the leading coefficient, and divide the angle by 3. This has actually a set of answers, since 0.pi., 2.pi. or 4.pi. can be added to that angle, before you divide it by 3 to take the cube root. So you end up with:

 

2^(1/6)e^(i.pi./12), 2^(1/6)e^(9i.pi./12) and 2^(1/6)e^(17.pi./12)

2^(1/6)e^(17.pi./12)