Rational power of complex number-calculation

[neginf]

Homework Statement

Use definition (1), Sec. 28 of z^c to show that (-1 + i*sqrt(3))^(3/2) = +/- 2*sqrt(2)

Homework Equations

z^c = e^(c*log z)

The Attempt at a Solution

(-1 + i * sqrt(3))^(3/2) = e^[(3/2) * log(-1 + i * sqrt(3))]
= e^[(3/2) * (ln| 4 | + i * arg(-1 + i * sqrt(3))]
= e^[(3/2) * (2 * ln 2 + i * 2* pi / 3)]
=e^[(3/2) * 2 * ln 2 + (3/2) * 2 * i * pi / 3 ]
=e^[3 * ln 2 + i * pi]
=e^[3 * ln 2] * e^(i * pi)
=(e^ln2)^3 * (cos(pi) + i * sin(pi))
=2^3 * (-1)
=-8.

 

[HallsofIvy]

That's not the way I would do it. I would use DeMoivre's formula:
$$(r e^{i\theta})^n= r^n e^{i n\theta}$$

Here, $z= -1+ i\sqrt{3}$. Drawing that point on the complex plane and dropping a perpendicular to the real axis gives a right triangle with legs of length 1 and $\sqrt{3}$. The hypotenuse has length $\sqrt{1^2+ \sqrt{3}^2}= \sqrt{1+ 3}= \sqrt{4}= 2$ so r= 2. Further, since the hypotenuse is exactly twice the leg along the real axis, duplicating the right triangle on the other side of the vertical line gives an equilateral triangle. The angle at the vertex is $\pi/3$ so $\theta= \pi- \pi/3= 2\pi/3$.

[tex]z= 2 e^{i2\pi/3}[itex]so $z^{3/2}= 2^{3/2}e^{(i2\pi/3)(3/2)}= 2^{3/2)e^{i\pi}$[/itex][/tex][itex][itex] reduce that to rectangular form.<br /> <br /> Of course, adding $2\pi$ to the argument will not change z but <br /> $$z^{3/2}= 2^{3/2}e^{(i(2pi/3+ 2\pi))(3/2)}= 2^{3/2}e^{(i 8\pi/3)(3/2)}= 2^{3/2}e^{i4\pi}$$[/itex][/itex]

 

[neginf]

Thank you for that. I see how you get the +/- 2*sqrt(2).
I think what I did wrong was the ln(|-1+i*sqrt(3)|) which is ln 2, not 2*ln 2 since |-1+i*sqrt(3)| is 2, not 4.