Trouble with complex numbers!!!?

Question

Solve the equotations please:

z*=3z and z+z*=1 (z* is the complex conjugate)

Answer 1

Sean H was OK with his analysis up until the very end. a=b=0 is perfectly acceptable; it just means that our complex number z = a + bi = 0 + 0i, which is fine. This is just the complex number 0!

For the second question, again we let z = x + yi. Then

z + z* = 1
x + yi + x - yi = 1
2x = 1
x = 1/2

Notice that the y's canceled out here. This means that ANY value of y will satisfy the given equation. So all we can say is that our complex number z must have real part equal to 1/2; the imaginary part can be any real number. Another way of saying this is that

z = 1/2 + yi

where y can be any real number.

Answer 2

say z = a + ib.

then z* = 3 z

means:

a-ib=3a+i3b

implies

a = 3 a and -b = 3 b

implies a = 0 and b = 0, so it's not possible to have z +z* = 1. Perhaps you copied the problem incorrectly?

Answer 3

if z= x+yi z*=x-yi
x-iy=3x+iy so 4x+2yi =0 x= 0 and y= 0
2x=1 and x= 1/2 y can be any value

Answer 4

Be a,b 2 real numbers such that z=a+bi
z*=a-bi

Plug into z*=3z

i) a-bi = 3(a+bi)

a-bi=3a + 3bi

Because a and b are real
a=3a -> a=0
-b = 3b -> b=0

z=0

ii) z+z*=1
a+bi + a - bi = 1

2a = 1

a=0.5

z=0.5 + bi for any real number b.