# High School Mathematics Extensions/Supplementary/Complex Numbers/Solutions

The imaginary unit

1. Compute $i^{25} = i$
2. Compute $i^{100} = 1$
3. Compute $i^{1000} = 1$
The pattern of $i^{1},i^{2},i^{3},...$ shows that $i^{4n} = 1$ where n is any integer. This case applies to questions 2 and 3. For question 1, $i^{25} = i^{24}\times i^{1} = i$.

Complex numbers as solutions to quadratic equations

1. Convince yourself that x = 3 - 2i is also a solution to the equation.
$\begin{matrix} x & = &3 - 2i &\\ x^2 & = & (3)^2 + 2(3)(-2i) + (-2i)^2 \\ & = & 5 - 12i\\ x^2 - 6x + 13 &=& 5 - 12i - 6(3-2i) + 13\\ &=& 0\\ \end{matrix}$
1. Plot the points A(3, 2) and B(3, -2) on a XY plane. Draw a line for each point joining them to the origin.
2. Compute the length of AO (the distance from point A to the Origin) and BO. Denote them by $r_1 \ \mbox{and} \ r_2$ respectively. What do you observe?
• $r_1 = \sqrt{3^2 + 2^2} = \sqrt{13}$
• $r_2 = \sqrt{3^2 + (-2)^2} = \sqrt{13}$
• $r_1 \ \mbox{and} \ r_2$ are the same
3. Compute the angle between each line and the x-axis and denote them by $\phi_1 \ \mbox{and} \ \phi_2$. What do you observe?
• $\phi_1 = \tan^{-1}(3/2) \approx 56$
• $\phi_2 = \tan^{-1}(3/-2) \approx -56$
• $\phi_1 \ \mbox{and} \ \phi_2$ differ only in the sign of the number
4. Consider the complex numbers:
$\begin{matrix} z & = & r_1 \cdot (\cos{\phi_1} + i\sin{\phi_1})\\ w & = & r_2 \cdot (\cos{\phi_2} + i\sin{\phi_2})\\ \end{matrix}$

Substitute z and w into the quadratic equation above using the values you have computed in Exercise 3 and 4. What do you observe? What conclusion can you draw from this?

$\begin{matrix} z & = & \sqrt{13} \cdot (\cos{\tan^{-1}(3/2)} + i\sin{\tan^{-1}(3/2)}) & = & 2+3i\\ w & = & \sqrt{13} \cdot (\cos{\tan^{-1}(3/-2)} + i\sin{\tan^{-1}(3/-2)})& = & 2-3i\\ \end{matrix}$

Thus the quadratic equation will equal 0 since z and w are equal to the solutions we found when solving the equation.

$\begin{matrix} x &=& 3 - 2i \\ y &=& 3 + 2i \end{matrix}$

Compute:

1. x + y
• (3 - 2i) + (3 + 2i) = 6
2. x - y
• 3 - 2i - (3 + 2i) = -4i
3. x2
• (3 - 2i)(3 - 2i) = 9 + (2)(3)(-2i) + 4i2 = 5 - 12 i
4. y2
• (3 + 2i)(3 + 2i) = 9 + (2)(3)(2i) + 4i2 = 5 + 12 i
5. xy
• (3 - 2i)(3 + 2i) = 9 + 6 i - 6i - 4i2 = 13
6. (x + y)(x - y)
• ((3 - 2i) + (3 + 2i))((3 - 2i) - (3 + 2i)) = (6)(-4i) = -24i

Division Convince yourself that the product of zw is always a real number. $zw = (a + bi)(a - bi) = a^2 - abi + abi -b^2i^2 = a^2 - b$

Complex Conjugate Convince yourself that the multiplication law is also true.

$\begin{matrix} z &= &a + bi\\ w &= &c - di\\ \\ x &=& z \times w \\ &=& (a + bi)(c - di) &=& ac + bci - adi - bdi^2 &=& ac - bd + (bc - ad)i \\ \bar{x}&=& ac - bd - (bc - ad)i \\ \\ \bar{z} \times \bar{w}& =& (a - bi)(c + di)\\ &=& ac - bic + adi - bdi^2\\ &=& ac - bd - (bc + ad)i \\ &=& \bar{x} \end{matrix}$

The complex root

1. Find (3 + 3i)1/2

$\begin{matrix} z & = & 3 + 3i \\ w & = & \sqrt{z}\\ w^2 & = & z\\ w &=& a + ib\\ w^2 &=& a^2 - b^2 + 2abi\\ \\ 3 &=& a^2 - b^2 &\mbox{(1)}&\\ 3 &=& 2ab &\mbox{(2)}&\\ \\ a &=& \frac{3}{2b} \\ \\ 3 &=& (\frac{3}{2b})^2 -b^2 \\ &=& \frac{9}{4b^2} -b^2 \\ 0 &=& \frac{9}{4b^2} -3 -b^2 \\ &=& \frac{9}{4} -3b^2 -b^4 \\ \\ c &=& b^2 \\ 0 &=& \frac{9}{4} -3c -c^2 \\ c &=& \frac{--3 \pm \sqrt{(-3)^2 -4\times \frac{9}{4} \times -1}}{2\times-1} \\ &=& \frac{-3 \pm \sqrt{18}}{2} \\ b^2 &=& \frac{-3 \pm \sqrt{18}}{2} \\ b &=& \sqrt{\frac{-3 +\sqrt{18}}{2}} \\ &=& \sqrt{\frac{-3 +\sqrt{18}}{2}} \\ &=& \sqrt{\frac{-3 +3\sqrt{2}}{2}} \\ &=& \sqrt{\frac{3(\sqrt{2}-1)}{2}} \\ &=& \sqrt{\frac{3}{2}}\sqrt{(\sqrt{2}-1)} \\ &=& \frac{\sqrt{6}}{2}\sqrt{(\sqrt{2}-1)} \\ &=& \frac{\sqrt{6(\sqrt{2}-1)}}{2}\\ \end{matrix}$
$\begin{matrix} 3 &=& 2ab\\ a &=& \frac{3}{2b} \\ &=& \frac{3}{2\frac{\sqrt{6(\sqrt{2}-1)}}{2}} \\ &=& \frac{3}{\sqrt{6}} \times \frac{1}{\sqrt{\sqrt{2}-1}} \\ &=& \frac{\sqrt{6}}{2} \times \frac{1}{\sqrt{\sqrt{2}-1}} \\ &=& \frac{\sqrt{6}}{2} \times \frac{\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1} \times \sqrt{\sqrt{2}+1}} \\ &=& \frac{\sqrt{6}}{2} \times \sqrt{\sqrt{2}+1} \\ &=& \frac{\sqrt{6(\sqrt{2}+1)}}{2} \end{matrix}$
Thus the solution for (3 + 3i)1/2 is:
$\sqrt{3+3i} = \pm (\frac{\sqrt{6(\sqrt{2}+1)}}{2} +\frac{\sqrt{6(\sqrt{2}-1)}}{2} \times i)$

2. Find (1 + 1i)1/2

$\begin{matrix} z & = & 1 + 1i \\ w & = & \sqrt{z}\\ w^2 & = & z\\ w &=& a + ib\\ w^2 &=& a^2 - b^2 + 2abi\\ \\ 1 &=& a^2 - b^2 &\mbox{(1)}&\\ 1 &=& 2ab &\mbox{(2)}&\\ \\ a &=& \frac{1}{2b} \\ \\ 1 &=& (\frac{1}{2b})^2 -b^2 \\ &=& \frac{1}{4b^2} -b^2 \\ 0 &=& \frac{1}{4b^2} -1 -b^2 \\ &=& \frac{1}{4} -b^2 -b^4 \\ \\ c &=& b^2 \\ 0 &=& \frac{1}{4} -1c -c^2 \\ c &=& \frac{--1 \pm \sqrt{(-1)^2 -4\times \frac{1}{4} \times -1}}{2\times-1} \\ &=& \frac{-1 \pm \sqrt{2}}{2} \\ b^2 &=& \frac{-1 \pm \sqrt{2}}{2} \\ b &=& \sqrt{\frac{-1 +\sqrt{2}}{2}} \\ &=& \frac{\sqrt{\sqrt{2} -1}}{\sqrt{2}} \\ &=& \frac{\sqrt{\sqrt{2} -1} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \\ &=& \frac{\sqrt{2(\sqrt{2} -1)}}{2} \\ \end{matrix}$
$\begin{matrix} 1 &=& 2ab\\ a &=& \frac{1}{2b} \\ &=& \frac{1}{2 \frac{\sqrt{2(\sqrt{2} -1)}}{2}} \\ &=& \frac{1}{\sqrt{2(\sqrt{2} -1)}} \\ &=& \frac{\sqrt{2(\sqrt{2} +1)}}{\sqrt{2(\sqrt{2} -1)} \times \sqrt{2(\sqrt{2} +1)}} \\ &=& \frac{\sqrt{2(\sqrt{2} +1)}}{1} \\ \end{matrix}$
Thus the solution for (1 + 1i)1/2 is:
$\sqrt{1+1i} = \pm (\frac{\sqrt{2(\sqrt{2}+1)}}{2} +\frac{\sqrt{2(\sqrt{2}-1)}}{2} \times i)$

3. Find i1/3

$\begin{matrix} z &=& i \\ w &=& z^{1/3} w^3 &=& z \\ \\ w &=& (a + ib)\\ w^3 &=& (a^2 - b^2 + 2abi) \times (a + ib)\\ z &=& (a^3 - 3ab^2) + i(3a^2b - b^3)\\ \\ 0 &=& a^3 - 3ab^2 \qquad (1)\\ 1 &=& 3a^2b - b^3 \qquad (2)\\ \end{matrix}$
$\begin{matrix} 0 &=& a^3 - 3ab^2\\ a^3 &=& 3ab^2 \\ a^2 &=& 3b^2 \\ \\ 1 &=& 3a^2b - b^3\\ 1 &=& 3(3b^2)b - b^3\\ 1 &=& 9b^3 - b^3\\ 1 &=& 8b^3\\ \frac{1}{8} &=& b^3 \\ b &=& \sqrt[3]{\frac{1}{8}} \\ &=& \frac{1}{2}\\ \\ a^2 &=& 3b^2 \\ &=& 3(\frac{1}{2})^2 \\ &=& \frac{3}{4} \\ a &=& \sqrt{\frac{3}{4}} \\ &=& \frac{\sqrt{3}}{2} \\ \end{matrix}$
Thus the solution for i1/3 is:

$\sqrt[3]{i} = \pm (\frac{\sqrt{3}}{2} + \frac{1}{2} \times i)$

Bases

1. The following numbers are written in base 2. Write them out in base 10:
1. 101011 (base 2) = 43 (base 10) = 25 + 23 + 21 + 20= 32+8+2+1
2. 001101 = 13
3. 10 = 2
4. 011 = 3
2. Write those numbers out in base 10 as if they were originally in base 5.
1. 101011 (base 5) = 3256 (base 10) = 55 + 53 + 51 + 50 = 3125+125+5+1
2. 001101 = 151
3. 10 = 5
4. 011 = 6
3. How many numbers could I write out in base 5 with only the first 4 columns?
Answer: 625 = 54 (each new column multiplies the number of possibilities by 5)
1. In computing, each 1 or 0 is called a bit. They are stored in groups of 8. Each group is called a byte. How many bytes are possible?
Answer: 256 = 28 (values from 00000000 through 11111111 binary, or 0 through 255 base 10)
1. Question: When editing the bytes directly, writing out 10110001 is too long and hexadecimal is used instead (in this case, B1). How many digits of hexadecimal are needed to cover all possible bytes?
Answer: 2 digits (162 = number of possible 2-digit hexadecimal numbers = 256 = number of possible bytes)