If \( z=(1+\sqrt{3}i)^{4/3} \), then the product of all the values of \( z \) is
Show Hint
For finding the product of all values of \( w^{1/n} \), write it as polynomial \( z^n - w = 0 \). The product of roots equals \( (-1)^n(-w) = w \) for odd \( n \).
Step 1: Turn the fractional power into a clean equation.
We have $z = (1+3i)^{4/3}$. To remove the fraction, cube both sides: \[ z^3 = (1+3i)^4 \] Step 2: Count the values of z.
The equation $z^3 = (1+3i)^4$ is a cubic, so it has exactly three values for $z$. Step 3: Use the product of roots idea.
Write it as $z^3 - (1+3i)^4 = 0$. For such an equation, the product of all three values equals the constant on top, which is $(1+3i)^4$. Step 4: Square the base first.
Compute $(1+3i)^2 = 1 + 6i + 9i^2 = 1 + 6i - 9 = -8 + 6i$. Step 5: Square again to reach the fourth power.
Now $(-8+6i)^2 = 64 - 96i + 36i^2 = 64 - 96i - 36 = 28 - 96i$. Step 6: State the product.
So the product of all the values of $z$ is \[ \boxed{28 - 96i} \]