SOLUTION: I am having trouble with these two questions. Can someone please explain. 1. Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbre

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I am having trouble with these two questions. Can someone please explain. 1. Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbre      Log On


   



Question 164201: I am having trouble with these two questions. Can someone please explain.
1. Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbreviated conj(z) or conj(a + bi) is the complex number a-bi. Also, the modulus of z, modulus(z) is the "size" of z, or SQRT(a^2 + b^2). Which of the following is true for all complex numbers? How did you get to that conclusion?
A) All of the following
B) z*conj(z) = [modulus(z)]^2
C) z + conj(z) = 2a
D) z - conj(z) = 2bi
E) None of the above

2. Using the definition of “size” of z from problem #1, which of the following is largest? How did you get to that conclusion?
A) 3i
B) -7
C) 4
D) -2i
E) 4 + 4i
F) None of the above

Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
A) All of the following
B) z*conj(z) = [modulus(z)]^2
C) z + conj(z) = 2a
D) z - conj(z) = 2bi
E) None of the above

B) right because (a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2=[modulus(z)]^2
C) right because (a+bi)+(a-bi)=2a
D) right because (a+bi)-(a-bi)=2bi

A) 3i size is sqrt(9)=3
B) -7 size is sqrt(49)=7
C) 4 size is sqrt(16)=4
D) -2i size is sqrt(4)=2
E) 4 + 4i size is sqrt(16+16)=sqrt(32)

so the largest is B)