A Complex Cofunction Equivalency For A Complex Exponential
The real number system, is algebraically incomplete. It cannot provide a solution to a simple equation like $x^2 = -1$. To solve this, we must define a new number, the imaginary unit i, with the fundamental property that $i^2 = -1$.
By this definition, $i$ is the square root of negative one. It cannot be expressed as a real number, as the square of any real number is always non-negative. This single extension to our number system unlocks a powerful new dimension of mathematics.
$i$ as orthogonal to $1$
The algebraic definition of i has a geometric implication.
If we consider the real numbers to lie on a one-dimensional number line, where does this new number i fit?
Multiplying by $-1$ corresponds to a $180^\circ$ rotation on this line, taking any value to its
additive inverse, $f:x\to i^2x=-x$.
Since $i^2 = -1$, multiplying by i twice is equivalent to a $180^\circ$ rotation. This implies that a single multiplication by i corresponds to a $90^\circ$ rotation.
Which is the concept of orthogonality. The imaginary axis is geometrically orthogonal (at a right angle) to the real axis, like the ordinate to the abscissa.
Euler's Formula provides the connection between the Cartesian and polar angle representations of the complex number, $z$. $z$ is equivalently expressed as an additive combination (like a binomial) of a real part, $\text{Re}(z)=x$, and a pure-imaginary part, $\text{Im}(z)=y\in \mathbb{R}$, appearing as $z=x+iy$. The
Taylor Series Proof of the Formula
The Tayler series of the exponential, with a complex argument, $z$, is equivalent to the sum of the Taylor of Cosine with an argment of $\text{Arg}(z)$ and imaginary component, Sine of the argument of $z$.
Euler's Formula states that for any real number $x$:
This can be shown using the Taylor series expansions for the exponential, sine, and cosine functions, and using the complex definition of $i$.
Logarithm of a Negative Number
$$ \log_2(-16) = \log_2(16) + \log_2(-1) = 4 + \frac{\ln(-1)}{ln(2)}= 4 + \frac{i\pi}{ln(2)} $$
The last step above is solved by using Euler's fomula for the imaginary exponent, $e^{ix} = \cos(x) + i\sin(x)$, which is derived using the Taylor series, in a subsequent article.
$$ e^{ix} = \cos(x) + i\sin(x) = -1 \rightarrow x=\pi + 2k\pi, k \in \mathbb{Z} \rightarrow \ln(-1)=i\pi $$
$$ y=2^x \rightarrow \ln(y) = x\ln(2) \rightarrow e^{\ln(y)} = e^{x\ln(2)} \rightarrow 2^x = e^{x\ln(2)} $$
$$ 2^{4+\frac{i\pi}{ln(2)}} = 16\left(2^{\frac{i\pi}{ln(2)}}\right) = 16e^{i\pi} = -16 $$
The Angle Addition Formula
The Angle Addition Formula states that for any angles $\theta$ and $\phi$:
Using Euler's formula, we can also express the sine and cosine functions in terms of complex exponentials:
$$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}, \quad \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$
Now, let's derive the angle addition formula for sine using these expressions:
$$ \sin(\theta + \phi) = \frac{e^{i(\theta + \phi)} - e^{-i(\theta + \phi)}}{2i} $$
We can expand the exponentials using the properties of exponents:
$$ = \frac{e^{i\theta}e^{i\phi} - e^{-i\theta}e^{-i\phi}}{2i} $$
Now, we can substitute the expressions for sine and cosine:
$$ = \frac{(\cos{\theta} + i\sin{\theta})(\cos{\phi} + i\sin{\phi}) - (\cos{\theta} - i\sin{\theta})(\cos{\phi} - i\sin{\phi})}{2i} $$
Factor out the terms:
$$ = \frac{(\cos{\theta}\cos{\phi} + i\sin{\theta}\cos{\phi} + i\sin{\phi}\cos{\theta} - \sin{\theta}\sin{\phi}) - (\cos{\theta}\cos{\phi} - i\sin{\theta}\cos{\phi} - i\sin{\phi}\cos{\theta} + \sin{\theta}\sin{\phi})}{2i} $$
Simplify the expression:
$$ = \frac{2i(\sin{\theta}\cos{\phi} + \cos{\theta}\sin{\phi})}{2i} $$
Finally, we arrive at the angle addition formula for sine:
$$ = \sin{\theta}\cos{\phi} + \cos{\theta}\sin{\phi} $$