Part I: Workout each of the following question with necessary steps clearly ad neatly 1. Use the principle of mathematical induction to prove that \( 6^{n}-1 \) is divisible by 5 , for all integers \( n \geq 1 \). ( 5 points) 2. If the complex number \( x+y i=\frac{c+i}{c-i} \), where \( x \) and \( y \) are real, prove that \( x^{2}+y^{2}=1 \) and \( \frac{y}{2 x}=\frac{c}{c^{2}-1} \). ( 5 points) 3. Find the fourth roots of a complex number \( z=-1+\sqrt{3} i \). 4. Show that the function \( f: Z \rightarrow Z \) given by \( f(n)=2 n+1 \) is not a one-to-one correspondence function. 5. Find the inverse of the function \( y=\log _{e} x+\sqrt{x^{2}+1} \) 6. Find the eccentricity, length of the transverse axis, length of the conjugate axis, vertices, foci, equations of the directrices, and length of the latus rectum of the hyperbola: ( 5 points ) (a) \( x^{2}-y^{2}=1 \) (b) \( 4 x^{2}-9 y^{2}=36 \)

1. To prove that \( 6^{n}-1 \) is divisible by 5 for all \( n \geq 1 \) using mathematical induction, we start with the base case \( n = 1 \): \( 6^1 - 1 = 5 \) which is divisible by 5. Now, assume it's true for some integer \( k \), i.e., \( 6^k - 1 \) is divisible by 5. For \( n = k + 1 \): \( 6^{k+1} - 1 = 6 \cdot 6^k - 1 = 5 \cdot 6^k + 6^k - 1 \). Since \( 5 \cdot 6^k \) is divisible by 5, we can conclude that \( 6^{k+1} - 1 \) is also divisible by 5. Therefore, by induction, \( 6^n - 1 \) is divisible by 5 for all \( n \geq 1 \. 2. Given \( x + yi = \frac{c+i}{c-i} \), we can multiply the numerator and denominator by the conjugate of the denominator: \( x + yi = \frac{(c+i)(c+i)}{(c-i)(c+i)} = \frac{c^2 + 2ci - 1}{c^2 + 1} \) thus, \( x = \frac{c^2 - 1}{c^2 + 1}, y = \frac{2c}{c^2 + 1} \). To show \( x^2 + y^2 = 1 \): Substitute \( x \) and \( y \) into the equation, simplify, and verify that it equals 1. Then, to prove \( \frac{y}{2x} = \frac{c}{c^{2}-1} \): Substitute \( x \) and \( y \) back into the equation and simplify to match the right side. 3. To find the fourth roots of the complex number \( z = -1 + \sqrt{3} i \), first express \( z \) in polar form: The modulus \( r = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2 \) and the argument \( \theta = \tan^{-1}(-\sqrt{3}/-1) = \frac{2\pi}{3} \). The fourth roots will then be given by \( \sqrt[4]{2} \text{cis} \left( \frac{2\pi/3 + 2k\pi}{4} \right) \) for \( k = 0, 1, 2, 3 \). Therefore, calculate to find the four unique root angles. 4. To show that \( f: \mathbb{Z} \rightarrow \mathbb{Z} \) given by \( f(n) = 2n + 1 \) is not one-to-one, assume \( f(a) = f(b) \). This means \( 2a + 1 = 2b + 1\). Subtracting gives \( 2a = 2b \) or \( a = b \), which confirms it is one-to-one. However, it is not a correspondence function as it does not cover all integers (e.g., \( f(n) \) only produces odd integers), lacking elements in the codomain (even integers) indicating it is not onto. 5. To find the inverse of \( y = \log_e x + \sqrt{x^2 + 1} \), first, isolate \( \log_e x \): Rearranging, you get: \( \log_e x = y - \sqrt{x^2 + 1} \), leading to \( x = e^{y - \sqrt{x^2 + 1}} \). Squaring both sides and solving for \( x \) yields the inverse function, \( x \) in terms of \( y \). 6. (a) For the hyperbola \( x^2 - y^2 = 1 \): - Eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{2} \) where \( a = 1 \) and \( b = 1 \). - Length of the transverse axis = 2a = 2, length of the conjugate axis = 2b = 2. - Vertices at \(