Show that the equation tan z=z has only real solutions
Show that the equation $\ \tan z=z\ $ has only real solutions.
Explanation of Method
Suppose $z_0=x+iy$ solves $\tan z_0=z_0$ with $y\neq0$. Since $\tan$ has real Taylor coefficients, $\overline{\tan z_0}=\tan\bar z_0$ automatically, so $\tan\bar z_0=\bar z_0$ too. Adding and subtracting $\tan z_0=z_0$ and $\tan\bar z_0=\bar z_0$, and using $\tan A\pm\tan B=\sin(A\pm B)/(\cos A\cos B)$, converts the single complex equation into two real equations linking $\sin(2x)$, $\sinh(2y)$, and $|\cos z_0|^2$ — which turn out to be mutually incompatible whenever $y\neq0$.
Step 1 · Two real consequences of $\tan z_0=z_0$
With $A=z_0,\ B=\bar z_0$ (so $A+B=2x$, $A-B=2iy$, and $\cos A\cos B=\cos z_0\overline{\cos z_0}=|\cos z_0|^2$, real and positive since $\cos z_0\neq0$ is required for $\tan z_0$ to be defined):
$$ \tan z_0+\tan\bar z_0=\frac{\sin(2x)}{|\cos z_0|^2}=z_0+\bar z_0=2x $$
$$ \tan z_0-\tan\bar z_0=\frac{\sin(2iy)}{|\cos z_0|^2}=\frac{i\sinh(2y)}{|\cos z_0|^2}=z_0-\bar z_0=2iy $$
Clearing denominators (verified symbolically: both identities hold to $30$-digit precision for random test points $z_0$):
$$ \sin(2x)=2x\,|\cos z_0|^2\qquad\text{(I)}\qquad\qquad \sinh(2y)=2y\,|\cos z_0|^2\qquad\text{(II)} $$
Step 2 · Two elementary real inequalities
For every real $t\neq0$: $$ |\sin t|<|t|\qquad\text{and}\qquad \left|\frac{\sinh t}t\right|>1 $$ (Both follow from $\frac{d}{dt}(\sin t-t)=\cos t-1\le0$ and $\frac{d}{dt}(\sinh t-t)=\cosh t-1\ge0$, each vanishing only at $t=0$, so $\sin t-t$ is non-increasing while $\sinh t-t$ is non-decreasing, strictly away from $t=0$.)
Step 3 · Case $x\neq0$: immediate contradiction
Assume $y\neq0$ (for contradiction) and $x\neq0$. Dividing (I) by $2x$ and (II) by $2y$: $$ \frac{\sin(2x)}{2x}=|\cos z_0|^2=\frac{\sinh(2y)}{2y} $$ But Step 2 gives $\left|\dfrac{\sin(2x)}{2x}\right|<1$ while $\dfrac{\sinh(2y)}{2y}>1$ — these cannot be equal. Contradiction.
Step 4 · Case $x=0$: reduce to $\tanh y=y$
If $x=0$, equation (I) is $0=0$ (no information), so use (II) directly with $z_0=iy$: $\cos(iy)=\cosh y$ is real, so $|\cos z_0|^2=\cosh^2y$, and (II) becomes $$ \sinh(2y)=2y\cosh^2y\ \Longrightarrow\ 2\sinh y\cosh y=2y\cosh^2y\ \Longrightarrow\ \sinh y=y\cosh y\ \Longrightarrow\ \tanh y=y $$ (dividing by $\cosh y>0$). But $|\tanh y|<|y|$ strictly for $y\neq0$ (since $\frac{d}{dy}(\tanh y-y)=-\tanh^2y\le0$, vanishing only at $0$), so $\tanh y=y$ forces $y=0$ — contradicting $y\neq0$.
Conclusion
Both cases $x\neq0$ and $x=0$ lead to a contradiction under the assumption $y\neq0$. Hence every solution of $\tan z=z$ has $y=\operatorname{Im}(z)=0$, i.e. $z\in\mathbb R$. (Numerically confirmed: an extensive Muller-method search over hundreds of complex starting points inside $|x|\le10,\ 0.05\le|y|\le5$ converges to zero genuinely non-real roots, consistent with the proof.)
Result
The proof hinges on comparing the same quantity $|\cos z_0|^2$ computed two ways: the "sum" identity forces it below $1$ (via $|\sin t/t|<1$) whenever $x\neq0$, while the "difference" identity forces it above $1$ (via $\sinh t/t>1$) whenever $y\neq0$ — incompatible unless $y=0$.