Expressing $e^z$ where $z=a+bi$ in polar form.
I am reading a passage of text that states:
"We can use the fact that $e^{a+bi}=e^a(\cos b+i\sin b)$ has polar form $\left<e^a,b \right>$ to verify that complex exponentials have various properties that we would expect, by analogy with real exponentials."
It then goes on to ask to express $e^z \times e^w$ in polar form where $z=a+bi$ and $w=c+di$.
Using the rule presented in the text above I can do this, what I am not clear on however is why this statement is true:
"We can use the fact that $e^{a+bi}=e^a(\cos b+i\sin b)$ has polar form $\left<e^a,b \right>$".
I expect I am overlooking something here but I am not making much progress. I am not sure how the polar form in this general case has been derived.
$\endgroup$3 Answers
$\begingroup$Let's do it piece by piece. In general, we know that the following is a property of the exponential function:
$$ e^{x + y} = e^x \cdot e^y $$
Replace $x$ with $a$ and $y$ with $b\,i$ to get:
$$ e^{a + b i} = e^a \cdot e^{b i} $$
Now for $e^{b\,i}$, use Euler's formula which states that:
$$ e^{b i} = \cos b + i\sin b $$
Apply this formula to get the result:
$$ e^{a + b i} = e^a (\cos b + i\sin b) $$
$\endgroup$ $\begingroup$By a polar form, $z=\rho e^{i\theta}$, for a complex number it is intedend the couple of the modulus $\rho$ and the phase $\theta$. So, $z=\langle \rho,\theta\rangle$. In your case is $\rho=e^a$ and $\theta=b$.
$\endgroup$ $\begingroup$For all $z\in\mathbb{C}$, we define:
$$ e^{z}=1+z+\frac{z^2}{2!}+...=\sum_{n=0}^{\infty}\frac{z^n}{n!} $$
Notice, that if we replace complex $z$ in real $x$, we'll get the expansion of the real value function $e^x$.
We prove that for all $z\in\mathbb{C}$ and $x\in\mathbb{R}$,
$$e^x\cdot e^z=e^{x+z}$$
By the above definition, we get: $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and $e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$. The series are converges absolutely, and therefor we can multiply them:
$$e^x\cdot e^z=1+(x+z)+(\frac{x^2}{2!}+xz+\frac{z^2}{2!})+...+(\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}z+...+\frac{z^n}{n!})+...=$$
$$=1+(x+z)+\frac{1}{2!}(x+z)^2+...+\frac{1}{n!}(x+z)^n+...=e^{x+z}$$
Now, let say that $x=a$ and $z=ib$, applying Euler's formula, we'll get out wanted result.
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