heureka

Let

\(z=-8+15i \text{ and }w=6-8i.\)

Compute

\(\dfrac{z\overline z}{w\overline w}\),

where the bar represents the complex conjugate.

\(\begin{array}{|rcl|rcl|} \hline z\overline z &=& |z|^2 & w\overline w &=& |w|^2 \\ \hline z &=&-8+15i & w&=&6-8i \\ |z| &=& \sqrt{8^2+15^2} & |w| &=& \sqrt{6^2+8^2} \\ |z|^2 &=& 8^2+15^2 & |w|^2 &=& 6^2+8^2 \\ \hline \mathbf{\dfrac{z\overline z}{w\overline w}} &=& \dfrac{|z|^2}{|w|^2} \\ &=& \dfrac{8^2+15^2}{6^2+8^2 } \\ &=& \mathbf{ \dfrac{289}{100} } \\ \hline \end{array} \)

There are two numbers whose 400th powers are equal to 9^1000.

In other words, there are two numbers that can replace x in the equation \(x^{400} = 9^{1000}\) making the equation true.
What are those numbers?

\(\begin{array}{|rcll|} \hline \mathbf{x^{400}} &=& \mathbf{9^{1000}} \quad | \quad \text{Take the 400th root of both sides } \\ x^{\frac{400}{400}} &=& 9^{\frac{1000}{400}} \\ x &=& 9^{\frac{5}{2}} \\ x &=& \sqrt{9^5} \\ \mathbf{ x} &=& \mathbf{\pm 243} \\ \hline \end{array}\)

1.
Let \(a_0=-2,b_0=1\), and for\( n\ge 0, let \begin{align*}a_{n+1}&=a_n+b_n+\sqrt{a_n^2+b_n^2},\\ b_{n+1}&=a_n+b_n-\sqrt{a_n^2+b_n^2}.\end{align*}\)
Find\( \dfrac{1}{a_{2012}} + \dfrac{1}{b_{2012}}\).

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{a_{n+1}} + \dfrac{1}{b_{n+1}}} &=& \dfrac{1}{a_n+b_n+\sqrt{a_n^2+b_n^2}} + \dfrac{1}{a_n+b_n-\sqrt{a_n^2+b_n^2}} \\\\ &=& \dfrac{a_n+b_n-\sqrt{a_n^2+b_n^2}+a_n+b_n-\sqrt{a_n^2+b_n^2}} {\left(a_n+b_n+\sqrt{a_n^2+b_n^2}\right)\left(a_n+b_n-\sqrt{a_n^2+b_n^2}\right)} \\\\ &=& \dfrac{2(a_n+b_n)} {\left(a_n+b_n\right)^2-\left(\sqrt{a_n^2+b_n^2}\right)^2} \\\\ &=& \dfrac{2(a_n+b_n)} {a_n^2+2a_nb_n+b_n^2-(a_n^2+b_n^2)} \\\\ &=& \dfrac{2(a_n+b_n)} { 2a_nb_n} \\\\ &=& \dfrac{ a_n+b_n } { a_nb_n} \\\\ &=& \mathbf{\dfrac{1}{a_{n}} + \dfrac{1}{b_{n}}} \\ \hline \mathbf{\dfrac{1}{a_{n+1}} + \dfrac{1}{b_{n+1}}} &=& \mathbf{\dfrac{1}{a_{n}} + \dfrac{1}{b_{n}}} = \ldots = \mathbf{\dfrac{1}{a_{1}} + \dfrac{1}{b_{1}}} \\\\ \dfrac{1}{a_{1}} + \dfrac{1}{b_{1}} &=& \dfrac{1}{-1+\sqrt{5}} + \dfrac{1}{-1-\sqrt{5}} = \dfrac{1}{2} \\\\ \hline \mathbf{\dfrac{1}{a_{2012}}+\dfrac{1}{b_{2012}}} &=& \dfrac{1}{a_{1}} + \dfrac{1}{b_{1}}\mathbf{ =\dfrac{1}{2}} \\ \hline \end{array}\)

Thank you, CPhill !

How many distinct ordered pairs of positive integers \((m,n)\) are there
so that the sum of the reciprocals of \(m\) and \(n\) is \(\dfrac14\)?

\(\text{From the relationship} \\ \dfrac{1}{z} = \dfrac{1}{m} + \dfrac{1}{n} \\ \text{follows immediately that $m>z$ and $n> z$ must be.}\\ \text{You can write $m=z+a$ and $n=z+b$ }\\ \text{Now the result:}\\ \dfrac{1}{z} = \dfrac{1}{z+a} + \dfrac{1}{z+b} \\ \)

\(\begin{array}{|rcll|} \hline \dfrac{1}{z} &=& \dfrac{1}{z+a} + \dfrac{1}{z+b} \\\\ \dfrac{1}{z} &=& \dfrac{2z+a+b}{z^2+za+zb+ab} \\\\ z^2+za+zb+ab &=& z(2z+a+b) \\ z^2+za+zb+ab &=& 2z^2+za+zb \\ z^2+za+zb+{\color{red}ab} &=& z^2+za+zb + {\color{red}z^2} \quad & \quad \text{by comparison follows } \boxed{z^2=ab} \\ \hline \end{array} \)

\(\text{Each pair $(a, b)=$ (divider, co-divider) of $n^2$ gives a solution }\\ \text{ from $\dfrac{1}{z} = \dfrac{1}{z+a} + \dfrac{1}{z+b} $.}\)

\(\text{if z = 4:}\\ \text{The divisors of $z^2=16$ are $1, 2, 4, 8, 16$ ($5$ divisors) }\)

\(\text{So there are $ \mathbf{5}$ distinct ordered pairs of positive integers $(m,n)$ }\)

\(\begin{array}{|c|c|c|c|c|} \hline 4^2 & divider & co-divider & \\ = 16 & a & b & ab & \dfrac{1}{4} = \dfrac{1}{4+a} + \dfrac{1}{4+b} \\ \hline & 1 & 16 & 1\cdot 16 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+1} + \dfrac{1}{4+16} = \mathbf{\dfrac{1}{5} + \dfrac{1}{20}} \\ \hline & 2 & 8 & 2\cdot 8 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+2} + \dfrac{1}{4+8}= \mathbf{\dfrac{1}{6} + \dfrac{1}{12}} \\ \hline & 4 & 4 & 4\cdot 4 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+4} + \dfrac{1}{4+4}= \mathbf{\dfrac{1}{8} + \dfrac{1}{8}} \\ \hline & 8 & 2 & 8\cdot 2 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+8} + \dfrac{1}{4+2}= \mathbf{\dfrac{1}{12} + \dfrac{1}{6}} \\ \hline & 16 & 1 & 16\cdot 1 = 16& \mathbf{\dfrac{1}{4} =} \dfrac{1}{4+16} + \dfrac{1}{4+1}= \mathbf{\dfrac{1}{20} + \dfrac{1}{5}} \\ \hline \end{array}\)

The distinct ordered pairs of positive integers \((m,n) = \sigma_0(z^2)\)

Let \(\large{a = 4 + 3i}\), \(\large{b = 1 -2i}\), and \(\large{c = 8 - 5i}\).  
The complex number d is such that a, b, c, and d form the vertices of a parallelogram,
when plotted in the complex plane. Enter all possible values of d, separated by commas.

\(\begin{array}{|rclclcl|} \hline \mathbf{d} &=& -a+b+c &=& -(4+3i)+ (1 -2i) + (8 - 5i) &=& \mathbf{5-10i} \\\\ \mathbf{d} &=& a-b+c &=& (4+3i)- (1 -2i) + (8 - 5i) &=& \mathbf{11+0i} \\\\ \mathbf{d} &=& a+b-c &=& (4+3i)+ (1 -2i) - (8 - 5i) &=& \mathbf{-3+6i} \\ \hline \end{array} \)

There are four complex numbers  such that
\(\large{z \overline{z}^3 + \overline{z} z^3 = 350}\),
and both the real and imaginary parts of  are integers.
These four complex numbers are plotted in the complex plane.
Find the area of the quadrilateral formed by the four complex numbers as vertices.

\(\text{Let $z=a+bi$}\\ \text{Let $\overline{z}=a-bi$}\)

\(\begin{array}{|rcll|} \hline \mathbf{z \overline{z}^3 + \overline{z} z^3} &=& \mathbf{350} \\\\ z \overline{z}\left( \overline{z}^2+z^2 \right) &=& 350 \\\\ && \boxed{z\overline{z} = |z|^2 =a^2+b^2\\ \overline{z}^2+z^2 =a^2-2abi-b^2 +a^2+2abi-b^2 =2(a^2-b^2)} \\\\ 2(a^2-b^2)(a^2+b^2) &=& 350 \\ (a^2-b^2)(a^2+b^2) &=& 175 \quad | \quad 175 = 7\cdot 25 \\ (a^2-b^2)(a^2+b^2) &=& 7\cdot 25 \\\\ \text{if } a>b \\ a^2-b^2 &=& 7 \quad (1) \\ a^2+b^2 &=& 25 \quad (2) \\ \hline (1)+(2): \quad 2a^2 &=& 32 \\ a^2 &=& 16 \\ \mathbf{ a } &=& \mathbf{\pm4} \\ \hline (2)-(1): \quad 2b^2 &=& 18 \\ b^2 &=& 9 \\ \mathbf{ b } &=& \mathbf{\pm3} \\ \hline \hline \end{array}\\ \begin{array}{|lcll|} \hline \text{The four complex numbers are: $\mathbf{(4+3i), (4-3i), (-4-3i), (-4+3i)} $} \\ \text{The four vertices are: $\mathbf{(4,3), (4,-3), (-4,-3), (-4,3)} $} \\ \text{The area of this rectangle is $8\cdot 6 = \mathbf{48} $} \\ \hline \end{array}\)

Evaluate

\(i^{11} + i^{16} + i^{21} + i^{26} + i^{31}\)

Formula: \(i^{4n+a} = i^a\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ i^{11} + i^{16} + i^{21} + i^{26} + i^{31} } \\ &=& i^{4\cdot 2+3} + i^{4\cdot 4+0} + i^{4\cdot 5+1} + i^{4\cdot 6+2} + i^{4\cdot 7+3} \\ &=& i^{3} + i^{0} + i^{1} + i^{2} + i^{3} \\ &=& i^{2}i + 1 + i + i^{2} + i^{2}i \quad | \quad i^2 = -1 \\ &=& -i + 1 + i -1 + -i \\ &=& \mathbf{ -i } \\ \hline \end{array}\)

Find

\( \left|\left(1 + \sqrt{3}i\right)^4\right|\)

\(\begin{array}{|rcll|} \hline \tan{60^\circ} &=& \dfrac{\sqrt{3}}{1} \\ |1 + \sqrt{3}i| &=& \sqrt{1^2 +\sqrt{3}^2 } \\ |1 + \sqrt{3}i| &=& 2 \\\\ 1 + \sqrt{3}i &=& 2\cdot e^{i\frac{60^\circ \pi}{180^\circ}} \\ (1 + \sqrt{3}i)^4 &=& 2^4\cdot e^{4i\frac{60^\circ \pi}{180^\circ}} \\ |(1 + \sqrt{3}i)^4| &=& 2^4 \\ \mathbf{|(1 + \sqrt{3}i)^4|} &=& \mathbf{16} \\ \hline \end{array}\)

For what value of  \(n\)  is  \(i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i\) ?

\(\begin{array}{|rcll|} \hline s &=& i + 2i^2 + 3i^3 + \cdots + ni^n \\ \hline \end{array}\)

\(\begin{array}{|rclll|} \hline s &=& i + 2i^2 + 3i^3 +4i^4+ \cdots + ni^n \\ is &=& \qquad i^2 +2i^3+3i^4+\ldots (n+1)i^n+ni^{n+1} \\ \hline s-is &=& \underbrace{ i+ i^2+i^3+i^4+\ldots +i^n}_{=S~ (GP)}-ni^{n+1}\\ s(1-i) &=& S-ni^{n+1}\\ &&\begin{array}{|rclll|} \hline S &=& i+ i^2+i^3+i^4+\ldots +i^n \\ iS &=& \qquad i^2+i^3+i^4+\ldots +i^n +i^{n+1} \\ \hline S - iS &=& i-i^{n+1} \\ S (1- i) &=& i( 1-i^{n}) \\ S &=& \dfrac{i( 1-i^{n})}{1-i} \\ \hline \end{array} \\\\ s(1-i) &=& \dfrac{i( 1-i^{n})}{1-i}-ni^{n+1} \\\\ s &=& \dfrac{i( 1-i^{n})}{(1-i)^2}-\dfrac{ni^{n+1}}{ 1-i } \\\\ s &=& \dfrac{i( 1-i^{n})}{(1-i)^2}-\left(\dfrac{ni^{n+1}}{ 1-i }\right)\cdot \left(\dfrac{1-i}{1-i}\right) \\\\ s &=& \dfrac{i( 1-i^{n}) -ni^{n+1}(1-i) }{(1-i)^2} \\\\ s &=& \dfrac{i( 1-i^{n}) -ni^{n}i(1-i) }{(1-i)^2} \\\\ s &=& \dfrac{i\Big( 1-i^{n} -ni^{n}(1-i)\Big) }{(1-i)^2} \quad | \quad (1-i)^2 = -2i \\\\ s &=& \dfrac{i\Big( 1-i^{n} -ni^{n}(1-i)\Big) }{-2i} \\\\ s &=& \dfrac{ 1-i^{n} -ni^{n}(1-i) }{-2 } \\\\ s &=& \dfrac{ -1+i^{n} +ni^{n}(1-i) }{2 } \\\\ s &=& \dfrac{ -1+i^{n} +ni^{n} -ni^{n+1} }{2 } \\\\ \mathbf{s} &=& \mathbf{\dfrac{ i^{n}(1+n) -i^{n+1}n -1 }{2 }} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{s} = \mathbf{\dfrac{ i^{n}(1+n) -i^{n+1}n -1 }{2 }} &=& 48 + 49i \\\\ i^{n}(1+n) -i^{n+1}n -1 &=& 96 + 98i \\\\ \mathbf{{\color{red}i^{n}}(1+n) -{\color{red}i^{n+1}}n } &=& \mathbf{97 + 98i} \\ \hline \end{array} \)

We compare the sides of the equation and note that i^n and i^(n+1) are adjacent values.

This means that one is either +i or -i and the other is either +1 or -1.

So there are two solutions for the comparison.

1. First possible solution

\(\begin{array}{|rcll|} \hline 98i &=&{ -\color{red}i^{n+1}}n \\ 97 &=& {\color{red}i^{n}}(1+n) \\ \hline \dfrac{98i}{97} &=& \dfrac{-i^{n+1} n}{i^{n} (1+n)} \\\\ \dfrac{98i}{97} &=& \dfrac{-i^{n}i\cdot n}{i^{n} (1+n)} \\\\ \dfrac{98i}{97} &=& \dfrac{- i\cdot n}{ (1+n)} \\\\ \dfrac{98}{97} &=& \dfrac{-n}{ (1+n)} \\\\ (1+n)98 &=& -97n \\ 98+98n &=& -97n \\ 195n &=& -98 \\ 195n &=& -98 \\ n &=& -\dfrac{98}{195} \\ n &=& -0.50256410256 \qquad \text{no solution, n is no integer } \\ \hline \end{array}\)

2. Second possible solution

\(\begin{array}{|rcll|} \hline 98i &=& {\color{red}i^{n}}(1+n) \\ 97 &=&{ -\color{red}i^{n+1}}n \\ \hline \dfrac{98i}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n+1} n} \\\\ \dfrac{98i}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n}i\cdot n} \\\\ \dfrac{98i^2}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n} \cdot n} \\\\ \dfrac{-98}{97} &=& \dfrac{1+n}{-n} \\\\ \dfrac{98}{97} &=& \dfrac{1+n}{n} \\\\ 98n &=& 97(1+n) \\ 98n &=& 97 + 97n \\ \mathbf{n} &=& \mathbf{97} \\ \hline \end{array}\)

\(i + 2i^2 + 3i^3 + \cdots + 97i^{97} = 48 + 49i\)