Let $$\theta_1, \theta_2, ......., \theta_{10}$$ be positive valued angles (in radian) such that $$\theta_1 + \theta_2 + ..... + \theta_{10} = 2 \pi$$. Define the complex numbers $$Z_1 = e^{i \theta_1}, Z_k = Z_{k - 1}e^{i \theta_k}$$ for $$k = 2, 3, ...., 10,$$ where $$i = \sqrt{-1}$$. Consider the statements P and Q given below:
$$P : \mid Z_2 - Z_1 \mid + \mid Z_3 - Z_2 \mid + .... + \mid Z_{10} - Z_9 \mid + \mid Z_1 - Z_{10} \mid \leq 2 \pi$$
$$Q : \mid Z_2^2 - Z_1^2 \mid + \mid Z_3^2 - Z_2^2 \mid + .... + \mid Z_{10}^2 - Z_9^2 \mid + \mid Z_1^2 - Z_{10}^2 \mid \leq 4 \pi$$
Then

We are given $$\theta_1,\theta_2,\ldots ,\theta_{10}\gt 0$$ with $$\theta_1+\theta_2+\cdots +\theta_{10}=2\pi$$ and the complex numbers

$$Z_1=e^{i\theta_1},\qquad Z_k=Z_{k-1}e^{i\theta_k}=e^{i(\theta_1+\theta_2+\cdots +\theta_k)}\; (k=2,3,\ldots ,10).$$

Thus the points $$Z_1,Z_2,\ldots ,Z_{10}$$ lie successively on the unit circle in the anticlockwise sense, and

$$Z_{10}=e^{i(\theta_1+\theta_2+\cdots +\theta_{10})}=e^{i\,2\pi}=1.$$

Case P : Evaluate $$S_1=\lvert Z_2-Z_1\rvert+\lvert Z_3-Z_2\rvert+\cdots +\lvert Z_{10}-Z_9\rvert+\lvert Z_1-Z_{10}\rvert.$$

For any two consecutive points $$e^{i\phi}$$ and $$e^{i(\phi+\theta_k)}$$ on the unit circle, the chord length is

$$\lvert e^{i\phi}-e^{i(\phi+\theta_k)}\rvert=2\sin\frac{\theta_k}{2}.$$

Therefore

$$S_1=\sum_{k=1}^{10}2\sin\frac{\theta_k}{2}.$$(1)

Using the inequality $$\sin x\le x$$ for $$x\gt 0$$, put $$x=\dfrac{\theta_k}{2}$$:

$$2\sin\frac{\theta_k}{2}\le 2\cdot\frac{\theta_k}{2}=\theta_k.$$(2)

Adding the ten inequalities in (2) and using $$\sum_{k=1}^{10}\theta_k=2\pi$$ gives

$$S_1\le 2\pi.$$(3)

Hence statement P is true.

Case Q : Evaluate $$S_2=\lvert Z_2^{\,2}-Z_1^{\,2}\rvert+\lvert Z_3^{\,2}-Z_2^{\,2}\rvert+\cdots +\lvert Z_{10}^{\,2}-Z_9^{\,2}\rvert+\lvert Z_1^{\,2}-Z_{10}^{\,2}\rvert.$$

The squared points are $$Z_k^{\,2}=e^{i2(\theta_1+\theta_2+\cdots +\theta_k)}$$ which are again on the unit circle (because modulus is 1). The argument difference between $$Z_{k-1}^{\,2}$$ and $$Z_k^{\,2}$$ is $$2\theta_k$$, so the corresponding chord length equals

$$\lvert Z_k^{\,2}-Z_{k-1}^{\,2}\rvert=2\sin\frac{2\theta_k}{2}=2\sin\theta_k.$$

Thus

$$S_2=\sum_{k=1}^{10}2\sin\theta_k.$$(4)

Apply again $$\sin x\le x$$ with $$x=\theta_k\;( \theta_k\gt 0):$$

$$2\sin\theta_k\le 2\theta_k.$$(5)

Summing (5) over $$k=1$$ to $$10$$ and using $$\sum_{k=1}^{10}\theta_k=2\pi$$ gives

$$S_2\le 2\cdot 2\pi=4\pi.$$(6)

Hence statement Q is also true.

Both P and Q are true, so the correct option is:

Option C which is: both P and Q are TRUE.