Evaluate (−1/2 + j1/2) ^−2/3

Question

Ans: 2^(1/3).e^(−jπ/2), 2^(1/3). e^(jπ/6), 2^(1/3). e^(j5/π)

Please show step by step.

Answer 1

So the complex number z = (-1/2, 1/2) [ where the second coordinate is implicitly multiplied by j ] raised to the exponent -2/3 can be seen from a different form...

z = re^jθ

Therefore, if we can represent (-1/2,1/2) into this form we can complete the problem...

r = magnitude(-1/2,1/2) = sqrt(1/4 + 1/4) = sqrt(2)/2

θ = arctan((1/2)/(-1/2)) = arctan(-1) = 135 deg = 3π/4

z = (-1/2, 1/2) = sqrt(2)/2 * e^(3πj/4)

z^(-2/3) = = 1/[z^{2/3)] = 1/[ r^(2/3)] * e^-jθ*2/3

(sqrt(2)/2)^2 = 2/4 = 1/2 --> (sqrt(2)/2)^(-2) = 2

(sqrt(2)/2)^(-2/3) = (2)^(1/3)

Now with the angular part... θ = 3π/4

e^-2jθ/3 = e^-j2*3π/12 --> e^-jπ/2

So the principal answer = 2^1/3 * e^-jπ/2

I you were to wrap around 2π from 3π/4 --> you would get 11π/4 and that would give another answer using θ = 11π/4. Likewise, "unrwrapping" by 2π from 3π/4 gives θ = -π/4.