Q13P - Mathematical Methods in Physical Sciences

Question

Express the following complex numbers in the $x+iy$ form. Try to visualize each complex number, using sketches as in the examples if necessary. The first twelve problems you should be able to do in your head (and maybe some of the others—try it!) Doing a problem quickly in your head saves time over using a computer. Remember that the point in doing problems like this is to gain skill in manipulating complex expressions, so a good study method is to do the problems by hand and use a computer to check your answers.

13. $\frac{{(i-\sqrt{3})}^3}{1-i}$

Step-by-Step Solution

Answer

The complex number in  form is  and the graph is shown below:

Step 1: Given information

The given complex number is $\frac{{(i-\sqrt{3})}^3}{1-i}$.

Step 2: Definition of Complex Number

The numbers that are presented in the form of a + ib, where, a,b  are real numbers and 'i' is an imaginary number called complex numbers.

Step 3: Find the value of z 1 and z 2

Let the given complex number be $\frac{{(i-\sqrt{3})}^3}{1-i}$.

 

Let z₁ and z₂ be 1 - i and $1-\sqrt{3}$ respectively.

 

The polar form of z₁  is $z_1=r_1{e}^{\left(i{\theta }_1\right)}$.

 

The polar form of z₂ is $z_2=r_2{e}^{\left(i{\theta }_2\right)}$.

 

Calculate the value of r₁, and ${\theta }_1$ as:

 

$\begin{array}{l}{r}_1=\sqrt{1+3}\\ r_1=2\end{array}$

 

And,

 

$\begin{array}{c}{\theta }_1=-\arctan \left(\frac{1}{\sqrt{3}}\right)\\ {\theta }_1=-\frac{\pi }{6}\\ {\theta }_1=\pi -\frac{\pi }{6}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} }\left\{\because {\theta }_1\text{ lies\hspace{0.17em} in\hspace{0.17em} I\hspace{0.17em} quadrant}\right\}\\ {\theta }_1=\frac{5\pi }{6}\end{array}$

 

Calculate the value of $r_2$ and ${\theta }_2$ as:

 

$\begin{array}{l}{r}_2=\sqrt{1+1}\\ r_2=\sqrt{2}\\ {\theta }_2=-\arctan \left(\frac{1}{1}\right)\\ {\theta }_2=-\frac{\pi }{4}\end{array}$

 

Solve further:

 

$\begin{array}{l}{\theta }_2=2\pi -\frac{\pi }{4}\left\{\because {\theta }_2\text{ lies\hspace{0.17em} in\hspace{0.17em} IV\hspace{0.17em} quadrant}\right\}\\ {\theta }_2=\frac{7\pi }{4}\end{array}$

 

The values of z₁ and z₂ are $2e^{(5\pi i/6)},\sqrt{2}{e}^{(7\pi i/4)}$ respectively.

Step 4: Find the value of z

Calculate the value of z as:

$\begin{array}{l}z=\frac{z_1}{z_2}\\ z=\frac{{\left[2e^{(5\pi i/6)}\right]}^3}{\sqrt{2}{e}^{(7\pi i/4)}}\\ z=4\sqrt{2}{e}^{\left[\right(5\pi i/2)-(7\pi i/4\left)\right]}\\ z=4\sqrt{2}{e}^{[3\pi i/4]}\end{array}$

The value of r and $\theta$ in z are $4\sqrt{2},\frac{3\pi }{4}$ respectively.

$\begin{array}{l}z=r\left(\cos \left(\theta \right)+i\sin \left(\theta \right)\right)\\ z=4\sqrt{2}\left(\cos \left(\frac{3\pi }{4}\right)+i\sin \left(\frac{3\pi }{4}\right)\right)\\ z=4\sqrt{2}\left(\left(\frac{-1}{\sqrt{2}}\right)+i\left(\frac{1}{\sqrt{2}}\right)\right)\\ z=-4+4i\end{array}$

The general form is $z=-4+4i$

Step 5: Draw the graph

Plot the complex number $z=-4+4i$.

 

 

Therefore, the general form is $z=-4+4i$.