Complex Numbers

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Enter the homework on complex numbers here.

Question 1 : Part (i)

Evaluate $(1+3i)+(5+7i)$

Add the real parts and add the imaginary parts:

$(1+5)+(3+7)i$

Simplify:

$6+10i$

Express Geometrically: When you add 2 complex numbers, you can think of vector addition.
The Answer is the diagonal of the parallelogram.
Hw1_Part1.bmp

++Question 1: Part (ii)

Evaluate $(1+3i)-(5+7i)$

Distribute the negative:

$1+3i-5-7i$

Add real and imaginary parts:

$(1-5)+(3-7)i$

Simplify:

$-4-4i$

Express Geometrically: When you subtract 2 complex numbers, you distribute the negative then think of vector addition.
Hw1_part2.bmp

Question 1 : Part (iii)

Evaluate $(1+3i)(5+7i)$

Use FOIL:

(1)
\begin{equation} 5+7i+15i-21 \end{equation}

Combine like terms:

(2)
\begin{equation} -16+22i \end{equation}

Question 1 : Part (v)

Evaluate $\sqrt{3+4i}$

Use the following formula for computing the roots of complex numbers

(3)
\begin{align} \sqrt[n]{z} = \sqrt[n]{r} \left[ \cos \left (\frac{\theta+2\pi k}{n} \right) + i \sin \left( \frac{\theta+2\pi k}{n} \right) \right], k=0,1,...,n-1 \end{align}

Determine the parameters to use in the formula

(4)
\begin{eqnarray} r & = & \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25} = 5 \\ \theta & = & \tan^{-1}(4/3) = 0.93 \\ n & = & 2 \\ k & = & 0,1 \end{eqnarray}

Substitute the parameters into the formula to obtain the values for the square root

(5)
\begin{align} (3+4i)^{1/2} = \sqrt[2]{5} \left[\cos \left(\frac{0.93-2\pi k}{2} \right)+i \sin \left(\frac{0.93-2\pi k}{2} \right) \right], k=0,1 \end{align}

Evaluate

(6)
\begin{align} \sqrt[2]{5} \left[\cos \left(\frac{0.93}{2} \right)+i \sin \left(\frac{0.93}{2} \right)\right] \approx 2 + i \end{align}
(7)
\begin{align} \sqrt[2]{5} \left[\cos \left(\frac{0.93-2\pi}{2} \right)+i \sin \left(\frac{0.93-2\pi}{2}\right)\right] \approx -2 - i \end{align}

Question 2:

Evaluate $\sqrt{1+i}$

$r = \sqrt[]{{1}^2+{1}^2}$

$r = \sqrt[]{2}$

$\Theta = tan^{-1}(\tfrac{1}{1})$

$\Theta = \tfrac{\pi}{4}$

$n = 2$

$k = 0,1$

$z = \sqrt[]{2}e^{{i}\tfrac{\pi}{4}}$

$z^\frac{1}{2} = (\sqrt[]{2}e^{{i}\tfrac{\pi}{4}})^\frac{1}{2}$

$k=0 \rightarrow \sqrt[4]{2}(cos(\frac{\pi}{8})+isin(\frac{\pi}{8}))$

$= \sqrt[4]{2}e^{{i}\tfrac{\pi}{8}}$

$=1.09868 + 0.45509 i$

$k=1 \rightarrow \sqrt[4]{2}(cos(\frac{\pi}{8} - \pi)+isin(\frac{\pi}{8} - \pi))$

$= - \sqrt[4]{2}e^{{i}\tfrac{\pi}{8}}$

$=- 1.09868 - 0.45509 i$

Question 3: Expand $\left ( cos\theta+isin\theta \right )^3$ $\; \; \; \; \; \; \; \; \; \;$ to obtain formulae for $cos3\theta$ $\; \;$ , $sin3\theta$ $\; \; \; \; \; \;$ in terms of $cos\theta$ $\; \;$ , $sin\theta$ $\; \; \;$ Use these formulae to show:

$cos3\theta=4cos^3\theta-3cos\theta$
$sin3\theta=3sin\theta-4sin^3\theta$

(8)
\begin{align} \left (cos\theta+isin\theta \right)^3=\left ( cos^2\theta+2icos{\theta}sin\theta+i^2sin^2\theta \right )\left ( cos\theta+isin\theta \right ) \end{align}
(9)
\begin{align} =cos^3\theta+2icos^2{\theta}sin\theta-sin^2{\theta}cos\theta+icos^2{\theta}sin\theta+2i^2cos{\theta}sin^2\theta+i^3sin^3\theta \end{align}
(10)
\begin{align} =cos^3\theta-isin^3\theta+3icos^2{\theta}sin\theta-3sin^2{\theta}cos\theta \end{align}

We can also equate $\left ( e^{i\theta} \right )^3$ to $\left ( cos\theta+isin\theta \right )^3$ and create a two-sided
equation:

(11)
\begin{align} e^{3i\theta}=cos^3\theta-isin^3\theta+3icos^2{\theta}sin\theta-3sin^2{\theta}cos\theta \end{align}

Now we group together the real and imaginary parts…

(12)
\begin{align} cos3\theta+isin3\theta=\left ( cos^3\theta-sin^2{\theta}cos\theta \right )+i\left ( -sin^3\theta+3cos^2{\theta}sin\theta \right ) \end{align}

…and create two separate equations

(13)
\begin{align} cos3\theta=cos^3\theta-3sin^2{\theta}cos\theta\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; sin3\theta=-sin^3\theta+3cos^2{\theta}sin\theta \end{align}

Working just with the $cos3\theta$ we get:

(14)
\begin{align} cos3\theta=cos^3\theta-3cos\theta\left ( 1-cos^2\theta \right ) \end{align}
(15)
\begin{align} =cos^3\theta-3cos\theta+3cos^3\theta \end{align}
(16)
\begin{align} =4cos^3\theta-3cos\theta \end{align}

And now with the $sin3\theta$ we get:

(17)
\begin{align} sin3\theta=-sin^3\theta+3sin\theta\left ( 1-sin^2\theta \right ) \end{align}
(18)
\begin{align} =-sin^3\theta+3sin\theta-3sin^3\theta \end{align}
(19)
\begin{align} =-4sin^3\theta+3sin\theta \end{align}

Question 5: Use the identities from Question 3 to show:

(20)
\begin{eqnarray} \cos^3 \theta & = & \frac{1}{4}(3 \cos\theta + \cos 3\theta ) \\ \sin^3 \theta & = &\frac{1}{4}(3 \sin\theta - \sin 3\theta) \end{eqnarray}

$cos3\theta = 4cos^3\theta -3cos\theta$
$4cos^3\theta = 3cos\theta +cos3\theta$
$cos^3\theta = 3cos\theta +cos3\theta * \frac{1}{4}$
$cos^3\theta = \frac{1}{4} (3cos\theta +cos3\theta)$

$sin3\Theta = 3sin\Theta - 4sin\Theta$
$4sin^3\Theta = 3sin\Theta - sin3\Theta$
$sin^3\Theta = 3sin\Theta - sin3\Theta * \frac{1}{4}$
$sin^3\Theta = \frac{1}{4} (3sin\Theta - sin3\Theta)$

Question 6: Use the trigonometric identities from Question 5 to evaluate

(21)
\begin{eqnarray} \int_{0}^{\frac{\pi}{2}} \cos^3\theta d\theta \\ \int_{0}^{\frac{\pi}{2}} \sin^3\theta d\theta \end{eqnarray}
(22)
\begin{eqnarray} \int_{0}^{\pi /2}cos^3\theta & = & \frac{1}{4} \int_{0}^{\pi /2} \left( 3 \cos \theta + \cos 3 \theta \right) d \theta \\ & = & \frac{1}{4} \left( 3 \sin \theta + \frac{1}{3} \sin 3 \theta \right) \left|_0^{\frac{\pi}{4}} \right. \\ & = & \frac{1}{4} \left( 3 - \frac{1}{3} \right) \\ & = & \frac{2}{3} \end{eqnarray}

$\int_{0}^{\pi /2}sin^3\theta =\int_{0}^{\pi /2}(1/4(3sin\theta +sin(3\theta ))d\theta$
$= 1/4[-3cos(\pi /2)+3cos(0)+(1/3)cos3[\pi /2]-cos3(0)]$
$= 1/4[-3(0)+3(1)+(1/3)(0) -1$]
$= 1/4[0+3-(1/3)]$
$= 1/4(2)$
$= (2/3)$
$\therefore \int_{0}^{\pi /2}sin^3\theta = (2/3)$

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