Recent content by hugo28

Thank you very much Vela. However, there is one small step left, would you please guide me. Thanks. (1-z)/(1+z) = 1 => (1-z)= (1+z) => 1-1= 0 <=> 2z=z+z, thus z=0 true. similar technique: (1-z)/(1+z)=w => (1-z)= w(1+z) => 0=w+zw-1+z => 0= (w-1)+z(w+1) Solve for z =-(w-1)/(w+1) => z=...

Revised: Divide both side by (1+z)^5, gives (1-z)^5 / (1+z)^5 = [(1-z) / (1+z)]^5=1 Let capital Z = [(1-z) /(1+z )], thus Z^5 = [(1-z) /(1+z)]^5 =1 Z^5 = 1 = (x + i*y) is (1+i*0y), r = (x^2 +y^2)^1/2 = (1^2 + 0)^1/2 = 1 (theta) = tan^(-1) |y/x| =0. z = r* e^(i*(theta)) [Z^5]^1/5 =...

Question: Given (-2*sqrt3) – ( 2*i) = 4*e^(i*7*pi/6) Perform the following: (a) Convention I: angle go from (0) to 2*pi. (b) Convention II: angle goes from (-)*pi to (+)*pi. = = = = = = = = = = = = = = = = = = = = = = = = = = = Part (a): Convention I: angle go from (0) to...

Divide both side by (1+z)^5: (1-z)^5 / (1+z)^5 = 1 Let Z1 = (1-z); (Z1)^5 = (1-z)^5 Z2 = (1+z); (Z2)^5 = (1+z)^5 Work numerator and denominator separately: (a) Numerator: (Z1)^5 = 1 = (cos(theta) + I sin(theta)) since (Z1)^5 = 1 => (x + iy) is...

Show/Prove that if f: X → Y is one-to-one on X, and A subset of X, then f^-1(f(A))<= subset A. If you wouldn’t mind, please check whether I did it correctly. Thanks in advance. Suppose x Є A Then, f(x) Є f(A) By image function y =f(x) Thus, y = f(x) Є f(A) And by inverse image...

Would you please help me? Thanks in advance. Prove that if a belongs to R, then (a^2)^1/2 = |a|? by using Abstract, Discrete, Algebraic Math. I work, but stuck: |a| =< (a^2)^1/2, then - (a^2)=< a =< (a^2) Please help!