Complex Numbers

[nycmathguy]

David Cohen

How about one more for tonight?

 

[MathLover1]

a.

let z=a+bi, w=c+d*i show that z¯=z

If z¯ be the conjugate of z then z¯ = z.

Proof:

Let z = a + bi where x and y are real and i = √-1.

Then by definition, (conjugate of z) = z¯ = a - ib.

Therefore, (conjugate of z¯) = z¯ = a + ib = z. Proved.

b.
____ _ _
(z+w)=z + w

Proof:
If z = a + ib and w = c + id then z¯ = a - ib and w¯ = c - id

Now, z - w = a + ib - c - id = a - c + i(b - d)

Therefore,
____ _ _ _ _
(z+w)=z + w = a - c - i(b - d)= a - ib - c + id = (a - ib) - (c - id) = z - w

 

[nycmathguy]

a.

let z=a+bi, w=c+d*i show that z¯=z

If z¯ be the conjugate of z then z¯ = z.

Proof:

Let z = a + bi where x and y are real and i = √-1.

Then by definition, (conjugate of z) = z¯ = a - ib.

Therefore, (conjugate of z¯) = z¯ = a + ib = z. Proved.

b.
____ _ _
(z+w)=z + w

Proof:
If z = a + ib and w = c + id then z¯ = a - ib and w¯ = c - id

Now, z - w = a + ib - c - id = a - c + i(b - d)

Therefore,
____ _ _ _ _
(z+w)=z + w = a - c - i(b - d)= a - ib - c + id = (a - ib) - (c - id) = z - w

I think there is a chapter on complex numbers just ahead in the Ron Larson textbook but I'm not sure. Nice work as always.