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.

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

dear! i am allowed to answer only one post. but you have posted too many. so, sorry. i am providing the first answer in this. please post the other questions in the next. thank you. Z is a vector space: (1) suppose a1+ib1, a2+ib2 are two complex numbers. then a1,a2,b1,b2 are real numbers. by Dedikind's axioms, we know that a1 + a2 and b1+b2 are also reals. so, (a1+a2) +i(b1+b2) is a complex number. ==> (a1+ib1)+(a2+ib2) is a complex number. thus, when ever a1+ib1, a2+ib2 are two complex numbers, (a1+ib1)+(a2+ib2) is a complex number. therefore, addition of complex numbers obeys closure law. -------------------------------------- we know by dedikind's axioms that addition of real numbers obeys associativity. i.e. a1+(a2+a3) = (a1+a2)+a3, b1+(b2+b3) = (b1+b2)+b3 ---(*) using these, we get (a1+ib1)+ {(a2+ib2)+(a3+ib3)}= (a1+ib1)+ {(a2+a3)i(b2+b3)} = {a1+(a2+a3)}+i{b1+(b2+b3)} ={(a1+a2)+a3}+i{(b1+b2)+b3} by(*) ={(a1+a2)+i(b1+b2)}+(a3+ib3) ={(a1+ib1)+(a2+ib2)}+(a3+ib3) thus, we have shown that X+(Y+Z) =(X+Y)+Z where X,Y,Z are complex numbers. so, addition of complex numbers obeys associativity. ---------------------------------------------------------------------------- note that we have to arrive from known to unknown. -------------------------------------------------------------------------- 0 is real number and so, 0+i0 is in Z such that (a+ib)+(0+i0) = (a+0)+i(b+0) by properties of real numbers, we know that a+0 = a, b+0 = b and so, the sum becomes a+ib. so, 0+i0 acts as the zero vector in Z. ----------------------------------------------------------------- suppose a+ib is a complex number. a, b are real -a and -b are real,. so, -a -ib is a complex number such that (a+ib)+(-a-ib) = (a-a)+i(b-b) = 0+i0 so, a+ib and -a-ib are the additive inverse vectors to each other in Z. --------------------------------------------------------------- (a+ib)+(c+id) = (a+c)+i(b+d) = (c+a)+i(d+b) by the abelian nature of addition of reals. =(c+id)+(a+ib) so, addition of complex numbers obeys abelian property. ------------------------------------------------------------------------------ suppose k is a scalar and a+ib is a vector. then a , b are also scalars. so, ka and kb are real. so, ka+i(kb) is a vector in Z. ==> k(a+ib) is a vector in Z. so, scalar multiplication of vector obeys closure law in Z ----------------------------------------------------------------------------------- suppose k,m are scalars and a+ib is a vector. k{ m(a+ib)} = k(ma+imb) ={k(ma)}+i{k(mb)} = {(km)a+i(km)b } =(km)(a+ib) ------------------------------------------------------------ (k+m)(a+ib)= (k+m)a+i(k+m)b = ( ka+ma)+i(kb+mb) = (ka+ikb)+(ma+imb) =k(a+ib)+m(a+ib) -------------------------------------------------------------- k{(a+ib)+(c+id)} = k{(a+c)+i(b+d)} = {k(a+c)+ik(b+d)} = (ka+kc)+i(kb+kd) = (ka+ikb) + (kc+ikd) = k(a+ib)+k(c+id) ---------------------------------------------------------- 1*(a+ib) = (1*a)+i(1*b) = a+ ib ------------------------------------------------------- thus, all the ten axioms of a vector space are satisfied. thus, Z forms a vector space over the field of reals.

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