Q1P - Mathematical Methods in Physical Sciences

Question

Show that if the line through the origin and the point z is rotated $\mathbf{90}\mathbf{°}$ about the origin, it becomes the line through the origin and the point iz. This fact is sometimes expressed by saying that multiplying a complex number by  rotates it through $\mathbf{90}\mathbf{°}$. Use this idea in the following problem. Let $\mathit{z}\mathbf{=}\mathit{a}{\mathit{e}}^{\mathbf{i}\mathbf{\omega }\mathbf{t}}$ be the displacement of a particle from the origin at time t. Show that the particle travels in a circle of radius a at velocity $\mathit{v}\mathbf{=}\mathit{a}\mathit{\omega }$ and with acceleration of magnitude directed toward the centre ${\mathit{v}}^{\mathbf{2}}\mathbf{/}\mathit{a}$ of the circle.

Step-by-Step Solution

Answer

It has been proved.

$$\begin{gathered} v=a\omega \\ A=a{\omega }^2 \end{gathered}$$

 

Step 1: Given Information.

The given expression is, $z=a{e}^{i\omega t}$.

Step 2: Meaning of rectangular form.

Represent the complex number in rectangular form means writing the given complex number in the form of x + iy in which x is the real part and y is the imaginary part.

Step 3: Change the angle and find equation.

Consider $z=e^{\left(i\theta \right)}$

Change the angle by adding $\frac{\pi }{2}$.

${\theta }_2=\theta +\frac{\pi }{2}$

Find the new number.

$$\begin{gathered} z=r{e}^{\left(i{\theta }_2\right)} \\ z=r{e}^{\left[\left(\theta +\pi /2\right)i\right]} \\ z=r{e}^{\left(\theta i\right)}.e^{\left(\pi i/2\right)} \\ z=r{e}^{\left(\theta i\right)}\left[\cos \left(\pi /2\right)+i \sin \left(\pi /2\right)\right] \\ z=ri{e}^{\left(\theta i\right)} \\ z=zi \end{gathered}$$

Step 4: Find the velocity.

Differentiate the equation with respect to time to find the velocity.

$$\begin{gathered} v=\frac{dz}{dt} \\ v=\frac{d}{dt}\left[a{e}^{\left(i\omega t\right)}\right] \\ v=ai\omega e^{\left(i\omega t\right)} \\ v=\omega i\left[a{e}^{\left(i\omega t\right)}\right] \\ v=\omega iz \end{gathered}$$

Find the magnitude.

$$\begin{gathered} \left|v\right|=\left|\omega iz\right| \\ \left|v\right|=\left|\omega i\right|.\left|z\right| \\ \left|v\right|=\omega a \end{gathered}$$

Step 5: Find the acceleration.

Differentiate the velocity with respect to time to find the acceleration.

$$\begin{gathered} v=\frac{dz}{dt} \\ v=\frac{d}{dt}\left[\omega ia exp\left(i\omega t\right)\right] \\ v=-a{\omega }^{2}exp\left(i\omega t\right) \\ v={\omega }^2\left[a exp\left(i\omega t\right)\right] \\ v={\omega }^{2}i \end{gathered}$$

Find the magnitude.

$$\begin{gathered} \left|A\right|=\left|{\omega }^{2}iz\right| \\ \left|A\right|=\left|\omega i\right|.\left|z\right| \\ \left|A\right|={\omega }^{2}a \end{gathered}$$

Therefore, it has been proved.

$$\begin{gathered} v=a\omega \\ A=a{\omega }^2 \end{gathered}$$