We are given the equation:

$z^{\frac{3}{2}} = 5i$

To solve for $z$, we need to isolate $z$ by raising both sides of the equation to the power of $\frac{2}{3}$. This will reverse the effect of raising $z$ to the power of $\frac{3}{2}$.

Step 1: Raise both sides to the power of $\frac{2}{3}$

$\left( z^{\frac{3}{2}} \right)^{\frac{2}{3}} = (5i)^{\frac{2}{3}}$

On the left-hand side, $\left( z^{\frac{3}{2}} \right)^{\frac{2}{3}}$ simplifies to $z$, so we have:

$z = (5i)^{\frac{2}{3}}$

Step 2: Simplify $(5i)^{\frac{2}{3}}$

To compute $(5i)^{\frac{2}{3}}$, it's helpful to express $5i$ in polar form. The complex number $5i$ has a magnitude of 5 and an argument of $\frac{\pi}{2}$ (since it's purely imaginary and points along the positive imaginary axis).

So, we can express $5i$ as: $5i = 5 \cdot e^{i\frac{\pi}{2}}$

Now, we compute $(5i)^{\frac{2}{3}}$ by raising the polar form to the power of $\frac{2}{3}$:

$(5i)^{\frac{2}{3}} = \left( 5 \cdot e^{i\frac{\pi}{2}} \right)^{\frac{2}{3}} = 5^{\frac{2}{3}} \cdot e^{i\frac{2}{3} \cdot \frac{\pi}{2}} = 5^{\frac{2}{3}} \cdot e^{i\frac{\pi}{3}}$

Step 3: Final Answer

Thus, the solution for $z$ is:

$z = 5^{\frac{2}{3}} \cdot e^{i\frac{\pi}{3}}$

This represents a complex number in polar form. We can convert this back into rectangular form (a + bi) if needed, but this is the general solution.

Would you like me to express it in rectangular form, or do you have any other questions about this solution?

Related Questions:

1. How do we convert complex numbers from polar to rectangular form?
2. What is the significance of the argument of a complex number in this type of problem?
3. How do we solve equations involving roots of complex numbers?
4. What is the geometric interpretation of raising a complex number to a fractional power?
5. How do we find all solutions to equations like this that involve complex numbers?

Tip: When dealing with complex numbers, the polar form is often the easiest way to work with powers and roots, especially when the argument is a simple fraction of $\pi$.