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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.

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up vote 3 down vote accepted

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:

$$ e^{a + b i} = e^a (\cos b + i\sin b) $$

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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 $\theta=b$.

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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:

$$e^x\cdot e^z=1+(x+z)+(\frac{x^2}{2!}+xz+\frac{z^2}{2!})+...+(\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}z+...+\frac{z^n}{n!})+...=$$


Now, let say that $x=a$ and $z=ib$, applying Euler's formula, we'll get out wanted result.

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