Find all complex number solutions for the equation \(z^4 = 16\).

Step 1 ：First, we can write the equation in the polar form. In polar form, a complex number \(z\) can be represented as \(z = r(cos(\theta) + isin(\theta))\), where \(r\) is the modulus of \(z\) and \(\theta\) is the argument of \(z\). In this case, \(r = \sqrt{16} = 4\) and \(\theta = 0\), so \(z = 4(cos(0) + isin(0))\).

Step 2 ：Next, we need to solve the equation \(z^4 = 16\). Using De Moivre's theorem, we know that if \(z = r(cos(\theta) + isin(\theta))\), then \(z^n = r^n(cos(n\theta) + isin(n\theta))\). Therefore, we get \(z^4 = 4^4(cos(4\times0) + isin(4\times0)) = 256\) which is not equal to 16. So, we need to find a new angle \(\theta\) such that \(z^4 = 16\). This gives us \(4^4(cos(4\theta) + isin(4\theta)) = 16\).

Step 3 ：Solving for \(\theta\) gives us \(\theta = \frac{k\pi}{2}\) for \(k = 0, 1, 2, 3\). Substituting these values back into the polar representation of \(z\) gives us the four complex roots of the equation \(z^4 = 16\).