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Let

$$z = e^{-\frac{a}{b + ic}}$$

(where $i$ is the imaginary unit and $a,b,c \in \mathbb{R}$) be a complex number in exponential form.

Write $z$ in the following form:

$$z = e^{- A}e^{-B}$$

with $A \in \mathbb{R} \setminus \{ 0 \}$, $B \in \mathbb{C}$.

I tried

$$\frac{a}{b + ic} \frac{b - ic}{b - ic} = \frac{a}{b^2 + c^2} (b - ic)$$

So

$$A = \frac{ab}{b^2 + c^2}, \ B = -\frac{iac}{b^2 + c^2}$$

Is this the only way to split the above exponent $a / (b + ic)$, or are there other ways to get a real $A$ and a purely imaginary (or complex) $B$?

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    $\begingroup$ You could split $\frac{a}{b^2 + c^2} (b - ic)$ into $\frac{ab}{b^2 + c^2} - \frac{ic}{b^2 + c^2}$, but apart from that there is not much you can do. $\endgroup$ – Tim The Enchanter Mar 14 '17 at 12:35
  • $\begingroup$ @TimTheEnchanter Yes, of course, I didn't write it but it was tacit. So, you think there is no other way to split the original $a / (b + ic)$? $\endgroup$ – BowPark Mar 14 '17 at 13:50
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    $\begingroup$ A pretty strange (or incomplete ?) question. $A=0,B=a/(b+ic)$ solves it. $\endgroup$ – Yves Daoust Mar 14 '17 at 14:22
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    $\begingroup$ @BowPark: not so helpful. Another solution can be $A=x,B=a/(b+ic)-x$. $\endgroup$ – Yves Daoust Mar 17 '17 at 10:05
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    $\begingroup$ @BowPark: as long as you allow $B$ complex, there is no unique solution. This was said in the accepted answer. $\endgroup$ – Yves Daoust Mar 17 '17 at 10:11
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Purely imaginary $B$

The real $A$ is unique. This is because when $z=\rho e^{i\theta}\in\mathbb{C}\backslash\{0\}$ where $\rho\in\mathbb{R}^+$ and $\theta\in\mathbb{R}$, $\rho$ is unique as $\rho=|z|$. To have the desired form, the only possibility for $A$ is such that $e^{-A}=\rho$, and that is $A=-\ln{\rho}=\ln{\frac{1}{\rho}}$.

$B$, on the other hand, is not unique. Recall that $$\forall\theta\in\mathbb{R},\,e^{i\theta}=\cos\theta+i\sin\theta$$ and since $\cos$ and $\sin$ are $2\pi$-periodic, we have $$\forall\theta\in\mathbb{R},\forall n\in\mathbb{Z},e^{i\theta}=e^{i(\theta+2\pi n)}$$

Thus, if $B$ is such that $z=e^{-A}e^{-B}$ then for any $n\in\mathbb{Z}$, $B_n=B+2\pi n$ satisfies $z=e^{-A}e^{-B_n}$ and $B_n$ is purely imaginary.

Complex $B$

When you don't ask for $B$ to be purely imaginary, even $A$ isn't unique:$$z=e^{-A}e^{-B}=e^{-\frac{A}{2}}e^{-\left(\frac{A}{2}+B\right)}=e^{-(A+B)}$$

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