# Question 8, Exercise 1.4

Solutions of Question 8 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.

Calculate the position of a particle from mean position when amplitude is $0.004 \mathrm{~mm}$ and angle is: $\dfrac{\pi}{4}$

Solution. Here we have $$x_{\max}=0.004, \quad \theta=\dfrac{\pi}{4}.$$ By using the formula \begin{align} x&=x_{\max} e^{i\theta} \\ &=0.004 e^{i\dfrac{\pi}{4}} \\ &=\frac{4}{1000} \left(\cos\left(\dfrac{\pi}{4}\right) +i \sin\left(\dfrac{\pi}{4}\right)\right) \\ &=\frac{1}{250} \left(\dfrac{1}{\sqrt{2}} +i \dfrac{1}{\sqrt{2}}\right) \\ &=\frac{1}{250\sqrt{2}}(1 +i) \\ &=\frac{\sqrt{2}}{500}(1 +i). \end{align} Hence mean position of particle is $\frac{\sqrt{2}}{500}(1 +i)$.

Calculate the position of a particle from mean position when amplitude is $0.004 \mathrm{~mm}$ and angle is: $\dfrac{\pi}{3}$

Solution.

$$x_{\max}=0.004, \quad \theta=\dfrac{\pi}{3}.$$ By using the formula \begin{align} x &= x_{\max} e^{i\theta} \\ &= 0.004 e^{i\dfrac{\pi}{3}} \\ &= \dfrac{4}{1000} (\cos(\dfrac{\pi}{3}) + i \sin(\dfrac{\pi}{3})) \\ &= \dfrac{4}{1000}(\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}) \\ &= \dfrac{1}{250} (\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}) \\ &= \dfrac{1}{500} + i \dfrac{\sqrt{3}}{500} \\ &= \dfrac{1}{500}\{1 + i\sqrt{3}\} \end{align} Hence the position of the particle from the mean position is $\dfrac{1}{500}\{1 + i\sqrt{3}\}$

Calculate the position of a particle from mean position when amplitude is $0.004 \mathrm{~mm}$ and angle is: $\dfrac{\pi}{6}$

Solution.

$$x_{\max}=0.004, \quad \theta=\dfrac{\pi}{6}.$$ By using the formula \begin{align} x &= x_{\max} e^{i\theta} \\ &= 0.004 e^{i\dfrac{\pi}{6}} \\ &= \dfrac{4}{1000} (\cos(\dfrac{\pi}{6}) + i \sin(\dfrac{\pi}{6})) \\ &= \dfrac{4}{1000}(\dfrac{\sqrt{3}}{2} + i \dfrac{1}{2}) \\ &= \dfrac{1}{250} (\dfrac{\sqrt{3}}{2} + i \dfrac{1}{2}) \\ &= \dfrac{\sqrt{3}}{500} + i \dfrac{1}{500} \\ &= \dfrac{1}{500}\{\sqrt{3} + i\} \end{align} Hence the position of the particle from the mean position is $\dfrac{1}{500}\{\sqrt{3} + i\}$