= Solution
For $v>0$ and $m>0$, let $\Phi$ and $\phi$ be the <standard normal distribution function> and density, and put
$$
d_1=\frac{-\log m+v/2}{\sqrt v},\qquad d_2=\frac{-\log m-v/2}{\sqrt v}=d_1-\sqrt v.
$$
The exercise event is $Z>(\log m+v/2)/\sqrt v=-d_2$. Complete the square in the first truncated expectation:
$$
\begin{aligned}
\mathbb E[e^{-v/2+\sqrt vZ}\mathbf1_{\{Z>-d_2\}}]
&=\int_{-d_2}^\infty e^{-v/2+\sqrt vz}\phi(z)dz\\
&=\int_{-d_2}^\infty\phi(z-\sqrt v)dz=\Phi(d_1).
\end{aligned}
$$
The probability of exercise is $\Phi(d_2)$. Hence the <normalized Black-Scholes call function> is
$$
\boxed{F(v,m)=\Phi(d_1)-m\Phi(d_2).}
$$
At $v=0$ it is $(1-m)^+$. If $m\le0$, the positive underlying always exceeds $m$, so $F(v,m)=1-m$; the reciprocal identity below is understood on its natural domain $m>0$.
For the reciprocal strike, $d_1(v,1/m)=-d_2(v,m)$ and $d_2(v,1/m)=-d_1(v,m)$. Thus
$$
\begin{aligned}
1-m+mF(v,1/m)
&=1-m+m\Phi(-d_2)-\Phi(-d_1)\\
&=1-m+m(1-\Phi(d_2))-(1-\Phi(d_1))\\
&=\Phi(d_1)-m\Phi(d_2)=F(v,m).
\end{aligned}
$$
Therefore
$$
\boxed{F(v,m)=1-m+mF(v,1/m),\qquad v\ge0,\ m>0.}
$$
The $v=0$ case follows either by continuity or by checking the two positive parts directly.
For the <stock> model, the original probability measure is already risk-neutral: $e^{-rt}S_t=S_0e^{\sigma W_t-\sigma^2t/2}$ is a true <martingale>. Write $\tau=T-t$. Conditional on $\mathcal F_t$, the <stock> at maturity is
$$
S_T=S_te^{r\tau}\exp\left(-\frac12\sigma^2\tau+\sigma\sqrt\tau Z\right),
$$
with $Z$ independent of current information and <standard normal distribution>. Therefore
$$
\mathbb E[e^{-r\tau}(S_T-K)^+\mid\mathcal F_t]=S_tF\left(\sigma^2\tau,\frac{Ke^{-r\tau}}{S_t}\right)=C_t(T,K).
$$
Equivalently,
$$
e^{-rt}C_t=\mathbb E[e^{-rT}(S_T-K)^+\mid\mathcal F_t],
$$
which is a true <martingale>, since the payoff is integrable. The sufficient direction of the <fundamental theorem of asset pricing> says that an equivalent measure making all discounted traded prices <local martingales> excludes <arbitrage> by <admissible trading strategies>. The original measure works simultaneously for the <bank account>, <stock> and call. Thus \b[the call-augmented market has no <arbitrage>].
For the put, hold one call, short one share, and hold $Ke^{-rT}$ units of the <bank account>. This <self-financing portfolio> has current value $C_t-S_t+Ke^{-r\tau}$ and terminal payoff
$$
(S_T-K)^+-S_T+K=(K-S_T)^+.
$$
Its discounted value is also the <conditional expectation> of that payoff. Thus the <put-call parity> price
$$
\boxed{P_t=Ke^{-r\tau}-S_t+C_t}
$$
adds another discounted <martingale> to the same market, proving \b[no <arbitrage> with both call and put included].
Now put $m=Ke^{-r\tau}/S_t$. The reciprocal identity gives
$$
P_t=S_t(F(\sigma^2\tau,m)-1+m)=S_tmF(\sigma^2\tau,1/m).
$$
The resulting <Put-call symmetry in the Black-Scholes model> is
$$
\boxed{P_t(T,K)=Ke^{-r(T-t)}F\left((T-t)\sigma^2,\frac{S_te^{r(T-t)}}K\right).}
$$
\b[The final formula printed in the PDF has two errors:] its discount-rate symbol is undefined, and the reciprocal argument has the wrong sign in the interest-rate exponential. Both are resolved by the displayed calculation. Even replacing the undefined symbol by $r$ leaves an incorrect expression with $e^{-r\tau}$ inside $F$. For $\tau>0$, $r>0$ and $\sigma>0$, $F(v,m)$ is strictly decreasing in $m$, because $\partial_mF=-\mathbb P(e^{-v/2+\sqrt vZ}>m)<0$. The printed argument is smaller than the reciprocal argument and hence gives a strictly larger put price. Thus the literal printed symmetry cannot be established; the boxed expression is the corrected identity.
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