Mathematics - Complex Number Question with Solution | TestHub

Given, ${\theta }_1+{\theta }_2+.....+{\theta }_{10}=2\pi$, $z_1=e^{i{\theta }_1}, z_k=z_{k-1}{e}^{i{\theta }_k}$

Now, $z_k=z_{k-1}{e}^{i{\theta }_k}$

$\Rightarrow z_2=z_1{e}^{i{\theta }_1}, z_3=z_2{e}^{i{\theta }_2}, \dots \dots$

As, $z_1=e^{i{\theta }_1}$$\Rightarrow \left|z_1\right|=1$

$z_2=z_1{e}^{{}_{i\theta }}$$\Rightarrow \left|z_2\right|=\left|z_1\right|=1$

Similarly $\left|z_1\right|=\left|z_2\right|\dots =\left|z_k\right|=1$

$\Rightarrow z_1,z_2,\dots ,z_k$ lies on a circle of unit radius.

We know, $\left|z_k-z_{k-1}\right|$ represents a line segment joining $z_k \& z_{k-1}$.

Both $z_k \& z_{k-1}$ lies on a unit circle

Since ${\theta }_1+{\theta }_2+....+{\theta }_{10}=2\pi$

We get 

So, $\left|z_2-z_1\right|,\left|z_3-z_2\right|,\dots ,\left|z_1-z_{10}\right|$ are the sides of a decagon circumscribed by a circle of unit radius

We know, the sum of the length of sides of a decagon is less than the circumference of its circumcircle.

$\Rightarrow \left|z_2-z_1\right|+\left|z_3-z_2\right|+\cdots +\left|z_{10}-z_9\right|+\left|z_1-z_{10}\right|\le 2\pi$

Hence, $P$ is true.

Similarly, ${\left(z_1\right)}^2=e^{i2{\theta }_1}, {\left(z_k\right)}^2={\left(z_{k-1}\right)}^2{e}^{i2{\theta }_k}, \dots \dots$

So, $\left|z_2^2-z_1^2\right|, \left|z_3^2-z_2^2\right|, \left|z_4^2-z_3^2\right|, \left|z_5^2-z_4^2\right|, \left|z_6^2-z_5^2\right|$ are 
the sides of a pentagon circumscribed by a circle of unit radius

So, $\left|z_2^2-z_1^2\right|+\left|z_3^2-z_2^2\right|+\dots \left|z_6^2-z_5^2\right|\le 2\pi$

Similarly, $\left|z_7^2-z_8^2\right|+\dots +\left|z_1^2-z_{10}^2\right|\le 2\pi$

Adding these equations, we get

$\left|z_2^2-z_1^2\right|+\left|z_3^2-z_2^2\right|+\cdots +\left|z_{10}^2-z_9^2\right|+\left|z_1^2-z_{10}^2\right|\le 4\pi$

Hence, $Q$ is true.