Write the payoffs in Figure 3.8 as linear combination of call options and derive a closed form formula for the Black-Scholes price, the Delta, and the Gamma of them. All the Greeks of the option are also linear combination of these call option Greeks. For instance, $$\Delta(t,S) = \Phi(d_1(\tau,K_1,S)) - \Phi(d_1(\tau,K_2,S)) - \Phi(d_1(\tau,K_3,S)) + \Phi(d_1(\tau,K_4,S))$$
Partial Solution: For the strangle we have a pay off of $$(K - S_T)_{+} + (S_T - K)_{+}$$ Therefore the closed form solution of B-S price of option is $$V(\tau,S) = P(\tau,K,S) + C(\tau,K,S)$$ and the delta of the position is $$\Delta(\tau,S) = -\Phi(-d_1(\tau,K,S)) + \Phi(d_1(\tau,K,S))$$ Finally our gamma for this position is $$\Gamma(\tau,S) = \frac{\Phi'(d_1(\tau,K,S)) + \Phi'(d_1(\tau,K,S))}{S\sigma \sqrt{\tau}}$$
I guess my professor made a mistake in regards to the B-S closed form price: for the strange it is $$V(\tau,S) = (-S_0\Phi(-d_1) + e^{-rT}K\Phi(-d_2)) + (S_0\Phi(d_1) - e^{-rT}K\Phi(d_2))$$ and similar for the straddle
where $\tau = T - t$ not sure why we use $\tau$ any explanation of that would be great.