Question

Let z1 = 2 − 2i and z2 = (1 − i) + √3(1 + i).

Answer 1

Answer:

Step-by-step explanation:

z₁ = 2 − 2i

z₂ = (1 − i) + √3(1 + i) = (1 + √3) + (√3 - 1) i

a) We get the modulus of z₁ as follows

║z₁║ = √((2)²+(-2)²) = 2

now we find the argument

α = Arctan (-2/2) = Arctan (-1) = -45º  ⇒   α = 360º + (-45º) = 315º

b) z₁ = 2 Cis 315º

Although the complex number is in binomic or polar form, its representation must be the same, since the complex number is the same, only that it is expressed in two different forms. The modulus represents the distance from the origin to the point. The degree of  rotation is the angle from the x-axis. When the polar form is expanded, the result is  the rectangular form of a complex number.

c) If  z₀*z₁ = z₂  and   z₀ = a + b i

we have

(a + b i)*(2 − 2i) = (1 + √3) + (√3 - 1) i

⇒  2a + 2bi - 2ai - 2bi² = (1 + √3) + (√3 - 1) i

⇒  2a + 2bi - 2ai - 2b(-1) = (1 + √3) + (√3 - 1) i

⇒  2a + 2b + 2bi - 2ai = (1 + √3) + (√3 - 1) i

⇒ 2 (a + b) + 2 (b - a) i = (1 + √3) + (√3 - 1) i

Now we can apply

2 (a + b) = 1 + √3

2 (b - a) = √3 - 1

Solving the system we get

a = 1/2

b = √3 / 2

Finally

z₀ = (1/2) + (√3 / 2) i

d) ║z₀║ = √((1/2)²+(√3 / 2)²) = 1

α = Arctan ((√3 / 2)/(1/2)) = 60º

e) z₀ = Cis 60º

f) Since z₂ = z₀*z₁, then z₂ is the transformation of z₁ rotated counterclockwise by arg(w) which is 60º