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

The imaginary unit
* Compute $i^{25}=i$
* Compute $i^{100}=1$
* 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
    ${\begin{matrix}x&=&3-2i&\\x^{2}&=&(3)^{2}+2(3)(-2i)+(-2i)^{2}\\&=&5-12i\\x^{2}-6x+13&=&5-12i-6(3-2i)+13\\&=&0\\\end{matrix}}$
* 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.
* 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
* 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
* 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.
Addition and multiplication
    ${\begin{matrix}x&=&3-2i\\y&=&3+2i\end{matrix}}$
Compute:
* x + y
** (3 - 2i) + (3 + 2i) = 6
* x - y
** 3 - 2i - (3 + 2i) = -4i
* x$^2$
** (3 - 2i)(3 - 2i) = 9 + (2)(3)(-2i) + 4i$^2$ = 5 - 12 i
* y$^2$
** (3 + 2i)(3 + 2i) = 9 + (2)(3)(2i) + 4i$^2$ = 5 + 12 i
* xy
** (3 - 2i)(3 + 2i) = 9 + 6 i - 6i - 4i$^2$ = 13
* (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. $zw=(a+bi)(a-bi)=a^{2}-abi+abi-b^{2}i^{2}=a^{2}-b$
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 i$^{1/3}$
    ${\begin{matrix}z&=&i\\w&=&z^{1/3}w^{3}&=&z\\\\w&=&(a+ib)\\w^{3}&=&(a^{2}-b^{2}+2abi)\times (a+ib)\\z&=&(a^{3}-3ab^{2})+i(3a^{2}b-b^{3})\\\\0&=&a^{3}-3ab^{2}\qquad (1)\\1&=&3a^{2}b-b^{3}\qquad (2)\\\end{matrix}}$
    ${\begin{matrix}0&=&a^{3}-3ab^{2}\\a^{3}&=&3ab^{2}\\a^{2}&=&3b^{2}\\\\1&=&3a^{2}b-b^{3}\\1&=&3(3b^{2})b-b^{3}\\1&=&9b^{3}-b^{3}\\1&=&8b^{3}\\{\frac {1}{8}}&=&b^{3}\\b&=&{\sqrt[{3}]{\frac {1}{8}}}\\&=&{\frac {1}{2}}\\\\a^{2}&=&3b^{2}\\&=&3({\frac {1}{2}})^{2}\\&=&{\frac {3}{4}}\\a&=&{\sqrt {\frac {3}{4}}}\\&=&{\frac {\sqrt {3}}{2}}\\\end{matrix}}$
    Thus the solution for i$^{1/3}$ is:
${\sqrt[{3}]{i}}=\pm ({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}\times i)$
Bases
* The following numbers are written in base 2. Write them out in base 10:
** 101011 (base 2) = 43 (base 10) = 2$^5$ + 2$^3$ + 2$^1$ + 2$^0$= 32+8+2+1
** 001101 = 13
** 10 = 2
** 011 = 3
* Write those numbers out in base 10 as if they were originally in base 5.
** 101011 (base 5) = 3256 (base 10) = 5$^5$ + 5$^3$ + 5$^1$ + 5$^0$ = 3125+125+5+1
** 001101 = 151
** 10 = 5
** 011 = 6
* How many numbers could I write out in base 5 with only the first 4 columns?
    Answer: 625 = 5$^4$ (each new column multiplies the number of possibilities by 5)
    Answer: 256 = 2$^8$ (values from 00000000 through 11111111 binary, or 0 through 255 base 10)
    Answer: 2 digits (16$^2$ = number of possible 2-digit hexadecimal numbers = 256 = number of possible bytes)