+73 1. Find the number of ordered pairs (a,b) of integers such that \(\frac{a + 2}{a + 5} = \frac{b}{3}\)
2. Find all complex numbers \(z\) such that \(z^4 = -4\)
Note: All solutions should be expressed in the form \(a+bi\), where \(a\) and \(b\) are real numbers.
1. There is only one solution: (a,b) = (4,2).
2. By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4). Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i. Then the other roots work out as
4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,
4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and
4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.
+26387 1.
Find the number of ordered pairs (a,b) of integers such that \(\mathbf{\dfrac{a + 2}{a + 5} = \dfrac{b}{3}}\)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a + 2}{a + 5}} &=& \mathbf{\dfrac{b}{3}} \\\\ 3(a+2) &=& b(a+5) \\ 3a+6 &=& ab+5b \\ -3a+5b+ab &=& 6 \\ -3a+5b+ab+(-3*5) &=& (-3*5)+6 \\ -3a+5b+ab-15 &=& -15+6 \\ \mathbf{(a+5)(b-3)} &=& \mathbf{-9} \\ \mathbf{m*n} &=& \mathbf{-9} \\ && \text{find all integer products }\ m*n = -9, \\ && \text{where}\ m=a+5,\ \text{ and }\ n=b-3 \\ \hline \end{array}\)
all integer products
\(\begin{array}{|ccl|r|r|r|r|} \hline && m*n & m & n & a=m-5 & b=n+3 \\ \hline -9 &=& 1\times -9 & 1 & -9 & -4 & -6 \\ -9 &=& -1\times 9 & -1 & 9 & -6 & 12 \\ -9 &=& 3\times -3 & 3 & -3 & -2 & 0 \\ -9 &=& -3\times 3 & -3 & 3 & -8 & 6 \\ -9 &=& 9\times -1 & 9 & -1 & 4 & 2 \\ -9 &=& -9\times 1 & -9 & 1 & -14 & 4 \\ \hline \end{array}\)
There are 6 integer products.
The number of ordered pairs (a,b) are 6.
\((-4,-6),\ (-6,12),\ (-2,0),\ (-8,6),\ (4,2),\ (-14,4)\)
+73 oh wait no its right, im so sorry. btw i really liked the way you explained it, i understand now! :)
