Complex number has the form : z=a+bi, where a and b are real numbers and i is th

Question

Complex number has the form : z=a+bi, where a and b are real numbers and i is the imaginary unit

(so, i=v-1. i2=-1). We define addition of complex numbers and scalar multiplication thusly:

(a1+b1i)+(a2+b2i)=(a1+a2)+(b1+b2)i

c(a+bi)=(ca)+(cb)i

a) Show that the set "Z" of all complex numbers forms a vector space with the given operations. (Finished)

b) Find a Basis, the dimension of the vector space "Z" , and the coordinates of z=3+4i

c) Find the matrix, relative to the Basis you gave in Part (b), for the "conjugate" Transformation

T: Z->Z that performs the operation T(a+bi)=a-bi. Show how this matrix computes the conjugate of the complex number z=3+4i

d) We define the inner product on Z: <a1+b1i,a2+b2i>=a1a2+b1b2. Use this inner product to find the distance and the angle between the vectors q=3+4i and w= -5+12i

Explanation / Answer

Basis: Let (a,b) represent a + bi. This clearly spans the entire space and can therefore be a valid basis as the two are linearly independent. Trivially, the coordinates of z would be (3,4). Conjugate Transpose: T * [a; b] = [a; -b] Let T = [1 0; 0 -1]. This matrix will map z = [a; b] to [a; -b]. [1 0; 0 -1] * [3; 4] = [3; -4]

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