Question 9, Exercise 1.4

Solutions of Question 9 of Exercise 1.4 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

When particle is at a position of $x=2+3 i$ from its mean position and $x_{\max }=1+4 i$ is the position at maximum distance from mean position as it can be seen under microscope at this point. Calculate the angle at time $\mathrm{t}=0$ and find the position of the particle.

Solution.

Here we have $$x=2+3i$$ $$x_{\max}=1+4 i$$ By using the formula $$\implies x=x_{\max} e^{i\theta}$$ $$2+3i=(1+4 i) e^{i\theta}$$ $$\begin{align} \implies e^{i\theta}&=\dfrac{2+3i}{1+4i} \\ &=\dfrac{(2+3i)(1-4i)}{(1+4i)(1-4i)} \\ &=\dfrac{2+12-6i+3i}{1+16} \\ &=\dfrac{14}{17}-\dfrac{5}{17}i. \end{align}$$

NOTE: This is not possible as $$|e^{i\theta}|=\left|\dfrac{14}{17}-\dfrac{5}{17}i \right| = \dfrac{\sqrt{221}}{17} \neq 1.$$ The contents, given in the textbook, related to these question are not suffient to solve such problems.

When particle is at a position of $x=2+3 i$ from its mean position and $x_{\max }=1+4 i$ is the position at maximum distance from mean position as it can be seen under microscope at this point. If $x=2+3 i$ and $x_{\max }=1+4 i$. Calculate the frequency when $\mathrm{t}=2$.

Solution.

The contents, given in the textbook, related to these question are not suffient to solve such problems.

< Question 8 Question 10 >