Solve $x^x=2x$ where $x\in\mathbb C$. 
Solve $x^x=2x$ where $x\in\mathbb C$.

Obviously, one solution is $x=2$. By WA, another solution is $x=0.346...$. How to solve it analytically, e.g. using Lambert W function?
Thank you.
 A: $$x^x=2x$$
$$=e^{x\ln x}=e^{(a+bi)(\ln|a+bi|+i(\arg(a+bi)+2k\pi))}$$
$$=e^{a\ln|a+bi|-b(\arg(a+bi)+2k\pi)}\cdot e^{bi\ln|a+bi|+ai(\arg(a+bi)+2k\pi)}$$
$$\implies {e^{a\ln|a+bi|}\over e^{b(\arg(a+bi)+2k\pi)}}=2|a+bi|$$
$$\text{and}\qquad e^{bi\ln|a+bi|+ai(\arg(a+bi)+2k\pi)}={a+bi\over|a+bi|}$$
$$\implies (a^2+b^2)^{a-1\over 2}=2e^{b\arg(a+bi)+2bk\pi}\tag 1$$
$$\text{and}\qquad\text{cis}(b\ln\sqrt{a^2+b^2}+a\arg(a+bi)+2ak\pi)=\text{cis}(\arg(a+bi)+2l\pi)\tag 2$$
$$(2)\implies \frac b2\ln(a^2+b^2)+(a-1)\arg(a+bi)=2l\pi-2ak\pi\tag 3$$
This is almost to the point of evaluating all-real numbers.  Fortunately, many graphing programs will evaluate $\arctan(\tfrac ba)$ using all branches and correctly evaluating at e.g. $x=0$ such that $\arg(a+bi)$ can be graphed as $\arctan(\tfrac ba)$.  Here are some graphs with $a$ along the $x$-axis and $b$ along the $y$-axis:


As demonstrated in the pictures, there are the three real solutions at $x\approx -0.658, 0.346, 2$, and there are many complex solutions, one of which is near the location $3+3i$.
Also, we have two equations to work with, so we can continue the direct analysis:
$$(1)\implies a^2+b^2=(2e^{b\arg(a+bi)+2bk\pi})^{2\over a-1}$$
Plugging into $(2)$:
$$\frac b2\ln((2e^{b\arg(a+bi)+2bk\pi})^{2\over a-1})+(a-1)\arg(a+bi)=2l\pi-2ak\pi$$
$$=\frac b{a-1}(b\arg(a+bi)+2bk\pi)\ln2+(a-1)\arg(a+bi)$$
$$=\frac {b^2\ln 2+(a-1)^2}{a-1}\arg(a+bi)+\frac{2b^2k\pi\ln 2}{a-1}$$
Unfortunately, this doesn't lend itself to any further breakdown that I am aware of.  The graph is not easy to create, since the branches appear to be a near-full covering of 2D space, and even setting e.g. $k=1$ only makes for confusing graphs that do not correspond to the solutions identified above.
