====== 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. 

=====Question 9(i)=====
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}

<fc #ff0000>NOTE: This is not possible as</fc> 
$$|e^{i\theta}|=\left|\dfrac{14}{17}-\dfrac{5}{17}i \right| = \dfrac{\sqrt{221}}{17} \neq 1.$$
<fc #ff0000>The contents, given in the textbook, related to these question are not suffient to solve such problems.</fc>


=====Question 9(ii)=====
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. **

<fc #ff0000>The contents, given in the textbook, related to these question are not suffient to solve such problems.</fc>





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