Find all complex roots of z^4 + 16 = 0 using De Moivre's Theorem.

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Step-by-Step Solution

Step 1: Rewrite the equation in polar form

First, we isolate $z^4$ by moving the constant term to the right side of the equation. Then, we express $-16$ in its polar form. The modulus is $16$, and since $-16$ lies on the negative real axis, its argument is $\pi$ radians. We add $2k\pi$ to account for all possible coterminal angles, where $k$ is an integer.

Step 2: Apply De Moivre's Theorem for roots

To find the roots, we take the fourth root of both sides of the equation. According to De Moivre's Theorem for roots, if $z = r(\cos\theta + i\sin\theta)$, then $z^{1/n} = r^{1/n}\left[\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right]$. Here, $r=16$, $\theta=\pi$, and $n=4.

Step 3: Calculate the modulus and simplify the argument

We calculate the fourth root of the modulus, $16^{1/4}$, which is $2$. Then, we simplify the argument by dividing each term by $4.

Step 4: Find the roots for k = 0, 1, 2, 3

We find the four distinct roots by substituting $k=0, 1, 2, 3$ into the simplified formula. For each value of $k$, we calculate the argument and then evaluate the cosine and sine values to express the root in rectangular form.