含虚数指数的多项式方程$b^{\pm i}-b = 3$的精确解法问询

Hey everyone, yesterday I came across one of those "99% of readers can't solve this" math problems that looked simple at first glance—but it led me to this equation with an imaginary unit in the exponent:

$$b^{\pm i}-b = 3.$$

The $\pm$ comes from a prior equation that had two distinct solutions, hence the two cases to consider.

I've managed to solve this numerically using Halley's method, but I'm hitting a wall trying to find exact solutions. I tried leveraging complex number properties to convert the base to $e$, switch to polar form, and split things into real/imaginary component systems, but I'm not sure where to go next. Here's what I've done so far:

Step 1: Convert to base $e$ to use Euler's formula #

My goal was to get a system of equations in the real domain, so first I rewrote the equation with base $e$:
$$e^{\pm i \log(b)} = 3+b.$$

Step 2: Expand the complex logarithm #

I used the complex logarithm definition (though I'm a bit rusty on the details here)—$\log(x) = \log(|x|)+i \cdot \arg(x)$—to avoid trig functions with complex angles:
$$e^{\pm i (\log(|b|)+i \cdot \arg(b))}=3+b.$$

This splits into two cases for the $\pm$:
\begin{equation}
\begin{cases}
e^{-\arg(b)+i \cdot \log(|b|)}=3+b,\
e^{\arg(b)-i \cdot \log(|b|)}=3+b.
\end{cases}
\end{equation}

Step 3: Apply Euler's formula #

Next, I expanded the left-hand sides using Euler's formula to separate real and imaginary parts:
\begin{equation}
\begin{cases}
e^{-\arg(b)}(\cos(\log(|b|)) + i \cdot \sin(\log(|b|)))=3+b,\
e^{\arg(b)}(\cos(\log(|b|)) - i \cdot \sin(\log(|b|)))=3+b.
\end{cases}
\end{equation}

Breaking those out fully:
\begin{equation}
\begin{cases}
e^{-\arg(b)}\cos(\log(|b|)) + ie^{-\arg(b)} \sin(\log(|b|))=3+b,\
e^{\arg(b)}\cos(\log(|b|)) - ie^{\arg(b)} \sin(\log(|b|))=3+b.
\end{cases}
\end{equation}

Step 4: Convert $b$ to polar form and split into real/imaginary systems #

I rewrote $b$ in polar form ($b = |b|\cos(\arg(b)) + i|b|\sin(\arg(b))$) and rearranged terms to set the real parts equal to 3 and imaginary parts equal to 0 (since the right-hand side has no imaginary component beyond $b$ itself):

For the first case ($-$ sign in the exponent):
\begin{equation}
\begin{cases}
e^{-\arg(b)}\cos(\log(|b|)) - |b|\cos(\arg(b)) = 3,\
e^{-\arg(b)}\sin(\log(|b|)) - |b|\sin(\arg(b)) = 0,
\end{cases}
\end{equation}

And for the second case ($+$ sign in the exponent):
\begin{equation}
\begin{cases}
e^{\arg(b)}\cos(\log(|b|)) - |b|\cos(\arg(b)) = 3,\
e^{\arg(b)}\sin(\log(|b|)) + |b|\sin(\arg(b)) = 0.
\end{cases}
\end{equation}

Where I'm stuck #

This is as far as I've gotten, and I'm not sure how to proceed from these real-valued systems to find exact solutions for $b$. I'm starting to wonder if exact solutions even exist, or if numerical methods are the only way here.

So my question is: Is there any way to solve this equation exactly, beyond numerical approximation?

备注：内容来源于stack exchange，提问作者Mowstyl