I need help with a worksheet

[Alice de]

These are the two exercises I need help with is:

1. Let x,y belong to R; if x-3+2i=3+y-yi, find the values of x and y.
2. Show that z+z*=2 Re(z) and z-z*=2i Im(z). What is z* equal to if z belongs to R? if z belongs to iR?

[mushroom]

Ok, so for the first question, you first need to realize that since you have real and imaginary components, the real components must be equal and the imaginary components must be equal.

This means 2i = -yi, meaning that y = -2

Through the same reasoning, we also know that x - 3 = 3 + y, and since y = -2, x - 3 = 3 - 2, and so x = 4

[IB Math Helper]

I'll answer both questions and I'll answer question 2 in parts (a) and (b) to represent the former and latter parts of the question.

1) By matching the imaginary parts of both sides of the equation, we see that 2i= -yi. So y=-2. If we match the real parts of both sides, then x-3=3+y. So x = 6+y = 6-2 = 4. Therefore, x=4 and y=-2.

2)

a) Let z = Re(z) + i Im(z). Correspondingly, its conjugate z* = Re(z) - i Im(z). So, z + z* = (Re(z) + i Im(z)) + (Re(z) - i Im(z)) = 2 Re(z). Similarly, z - z* = (Re(z) + i Im(z)) - (Re(z) - i Im(z)) = Re(z) + i Im(z) - Re(z) + i Im(z) = 2 i Im(z).

b) If z belongs to R, then Im(z) = 0. So z = Re(z) + i Im(z) = Re(z) + i (0) = Re(z). Since z* = Re(z) - i Im(z), then z* = Re(z) - i (0) = Re(z).

Similarly, if z has no real component, then Re(z) = 0. So, z = Re(z) + i Im(z) = 0 + Im(z) = Im(z). Since z* = Re(z) - i Im(z), then z* = 0 - Im(z) = - Im(z).