Q3P - Mathematical Methods in Physical Sciences

Question

For each of the following numbers, first visualize where it is in the complex plane. With a little practice you can quickly find $\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{r}\mathbf{,}\mathit{\theta }$ in your head for these simple problems. Then plot the number and label it in five ways as in Figure 3.3. Also, plot the complex conjugate of the number $\mathbf{1}\mathbf{-}\mathit{I}\sqrt{\mathbf{3}}$.

Step-by-Step Solution

Answer

The right values for the complex numbers are

$$\begin{gathered} x=1,y=-\sqrt{3} \\ r=2 \\ \theta =\frac{5\mathrm{\pi }}{3} \\ 1-i\sqrt{3}=2\left(\cos \frac{5\mathrm{\pi }}{3}+i \sin \frac{5\mathrm{\pi }}{3}\right) \\ =2e^{i\frac{5\mathrm{\pi }}{3}} \end{gathered}$$

Step 1: Given information and the general form of complex numbers:

The given complex number is $1-i\sqrt{3}$.

The general form of complex numbers is x+iy  .

Step 2: Complex numbers and the given conjugate:

The general form of complex numbers is x+iy .

$$\begin{gathered} x+iy=r\left(\cos \theta +i \sin \theta \right) \\ =r{e}^{i\theta } \\ r=\sqrt{x^2+y^2} \\ \theta ={\sin }^{-1}\frac{y}{r} \end{gathered}$$ 

 

In the given number$1-i\sqrt{3}$.

$$\begin{gathered} x=1,y=-\sqrt{3} \\ r=\sqrt{1^2+{\left(-\sqrt{3}\right)}^2} \\ =2 \\ \theta ={\sin }^{-1}\frac{-\sqrt{3}}{2} \\ =\frac{5\mathrm{\pi }}{3} \\ 1-i\sqrt{3}=2\left(\cos \frac{5\mathrm{\pi }}{3}+i \sin \frac{5\mathrm{\pi }}{3}\right) \\ =2e^{i\frac{5\mathrm{\pi }}{3}} \end{gathered}$$

Plot the given complex number and complex conjugate of the number on the graph.