Find all complex numbers $z$ such that $z^4 = -4.$
Note: All solutions should be expressed in the form $a+bi$, where $a$ and $b$ are real numbers.
To find the complex numbers $z$ that satisfy $z^4 = -4$, we can start by expressing $-4$ in its trigonometric form.
The modulus of $-4$ is $| -4 | = 4$. The argument of $-4$ is $\arg(-4) = \pi$.
In general, any complex number $z$ can be expressed in its trigonometric or polar form as $z = r(\cos \theta + i \sin \theta)$, where $r = |z|$ and $\theta = \arg(z)$.
Thus, we can write $-4$ as $-4 = 4 \left( \cos(\pi) + i \sin(\pi) \right)$.
Now, let's express $z^4$ using its trigonometric form:
$z^4 = \left( r(\cos \theta + i \sin \theta) \right)^4$
Using De Moivre's Theorem, we can expand this expression as follows:
$z^4 = r^4 (\cos(4\theta) + i \sin(4\theta))$
For $z^4$ to be equal to $-4$, we must have:
$r^4 (\cos(4\theta) + i \sin(4\theta)) = 4 \left( \cos(\pi) + i \sin(\pi) \right)$
This leads to two equations:
$r^4 = 4$
$\cos(4\theta) = \cos(\pi)$
Solving the first equation for $r$, we find:
$r = \sqrt[4]{4} = \sqrt{2}$
Solving the second equation for $\theta$, we find:
$4\theta = \pi + 2k\pi$ (where $k$ is an integer)
$\theta = \frac{\pi}{4} + \frac{k\pi}{2}$
Therefore, the solutions for $z$ are given by:
$z = \sqrt{2} \left( \cos \left( \frac{\pi}{4} + \frac{k\pi}{2} \right) + i \sin \left( \frac{\pi}{4} + \frac{k\pi}{2} \right) \right)$
where $k$ can be any integer.
To find all the complex numbers $z$ such that $z^4 = -4$, we can start by expressing $-4$ in exponential form.
We have $-4 = 4 \cdot e^{i\pi}$, where $e^{i\pi}$ is Euler's formula: $e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1$.
So, we can rewrite $z^4$ as $4 \cdot e^{i\pi}$.
Now, let's express $z$ in exponential form as well: $z = re^{i\theta}$, where $r$ is the magnitude of $z$ and $\theta$ is the argument of $z$.
Substituting this into the equation $z^4 = 4 \cdot e^{i\pi}$, we get $(re^{i\theta})^4 = 4 \cdot e^{i\pi}$.
Using the properties of exponents, we can rewrite this as $r^4 \cdot e^{4i\theta} = 4 \cdot e^{i\pi}$.
Comparing the magnitudes on both sides of the equation, we have $r^4 = 4$.
Taking the fourth root of both sides, we get $r = \sqrt[4]{4} = \pm \sqrt{2}$.
Now, let's look at the arguments. We have $e^{4i\theta} = e^{i\pi}$.
Using Euler's formula, we know that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$.
So, $e^{4i\theta} = \cos(4\theta) + i\sin(4\theta)$.
Comparing the real parts of both sides, we have $\cos(4\theta) = -1$.
Since the cosine function repeats every $2\pi$, we know that $4\theta = \pi + 2n\pi$, where $n$ is an integer.
Simplifying this equation, we get $\theta = \frac{\pi}{4} + \frac{n\pi}{2}$.
Now, let's combine the values we found for $r$ and $\theta$.
For $r = \sqrt{2}$ and $\theta = \frac{\pi}{4} + n\frac{\pi}{2}$, the complex numbers $z$ are given by:
$z = \sqrt{2} \cdot e^{i(\frac{\pi}{4} + \frac{n\pi}{2})}$.
Simplifying further, we have:
$z = \sqrt{2} \cdot \left(\cos\left(\frac{\pi}{4} + \frac{n\pi}{2}\right) + i\sin\left(\frac{\pi}{4} + \frac{n\pi}{2}\right) \right)$.
Expanding the trigonometric functions using their sum formulas, we find:
$z = \sqrt{2} \cdot \left(\frac{1}{\sqrt{2}}\cos\left(\frac{n\pi}{2}\right) - \frac{1}{\sqrt{2}}\sin\left(\frac{n\pi}{2}\right) + i\left(\frac{1}{\sqrt{2}}\cos\left(\frac{n\pi}{2}\right) + \frac{1}{\sqrt{2}}\sin\left(\frac{n\pi}{2}\right)\right) \right)$.
Simplifying this expression, we finally get:
$z = \frac{1}{\sqrt{2}}\left(\cos\left(\frac{n\pi}{2}\right) - \sin\left(\frac{n\pi}{2}\right) + i\cos\left(\frac{n\pi}{2}\right) + \sin\left(\frac{n\pi}{2}\right) \right)$.
Therefore, the complex numbers $z$ that satisfy the equation $z^4 = -4$ are given by $z = \frac{1}{\sqrt{2}}\left(\cos\left(\frac{n\pi}{2}\right) + i\sin\left(\frac{n\pi}{2}\right) \right)$ for $n \in \mathbb{Z}$.
-4 = 4cisπ
so, if z^4 = -4,
z = √2 cis(π/4 + kπ/2) for k=0,1,2,3