1. Given the complex numbers z1 = 2 − i and z2 = 1 + i.?

Question

1. Given the complex numbers z1 = 2 − i and z2 = 1 + i.
(a) Find z3 = z1 z2.
(b) Find z4 = z1/z2
.(note zk is z subscript k and rk is r subscript k)
(c) Express each zk in polar form zk = rk (cos θk + i sin θk)
; − π< θk ≤ π for k = 1, 2, 3 and 4.
(d) (i) Show r3 = r1 r2.
(ii) Show r4 = r1/r2
.
(iii) Show θ3 = θ1 + θ2.
(iv) Show θ4 = θ1 − θ2.
(e) Plot the complex numbers zk for k = 1, 2, 3 and 4 in the complex plane.

Answer 1

z1 = 2 − i
z2 = 1 + i
z3= 2*1+2i -i -i^2= 3+i
z4 = (2-i)(1-i)/(1+i)^2 = (1-3i)/(2)

You should be able to work out the rest.
r^2= x^2+y^2
tan(theta) = y/x
r = |x+yi| sqrt(x^2+y^2)

Answer 2

i did no longer understand what you have been asking in spite of the incontrovertible fact that it truly is log(z1z2) = logz1 + log z2 for all z1, z2 greater effective than 0. yet i think of log(z1z2) = logz1 + logz2 + 2n(pi)r for all nonzero complicated numbers. occasion: log a million = log1 + log1, a million =1cis90 the place r =a million. log1 = logz1 + logz2 + 2(0)pi(a million) = log1 + log1

Answer 3

Use properties of complex numbers to solve this problem.
(a) z3=(2-i)(1+i)=3+i
(b) z4=(2-i)/(1+i)=(1/2)(1-3i)
(c) Find the absolute value of the complex vector and the angle.
The rest of the problems should be easy.
Give it a try......