Question

Prove that Re(z1,z2-bar) = modulus of z1 * modulus of z2 iff arg z1 = arg z2 + 2*pi*n.Use polar form with arg measured in radians​

Answer 1

Explanation:

We can start by writing the given equation in terms of polar form:

Re(z1,z2-bar) = modulus of z1 * modulus of z2

If we write z1 and z2 in polar form, we get:

z1 = r1(cosθ1 + i sinθ1)

z2 = r2(cosθ2 + i sinθ2)

Then, the conjugate of z2 is:

z2-bar = r2(cosθ2 - i sinθ2)

Using the formula for the real part of the product of two complex numbers, we get:

Re(z1,z2-bar) = Re(z1 * z2-bar)

Substituting the expressions for z1 and z2-bar, we get:

Re(z1,z2-bar) = Re(r1(cosθ1 + i sinθ1) * r2(cosθ2 - i sinθ2))

Simplifying the product, we get:

Re(z1,z2-bar) = Re(r1r2[(cosθ1 cosθ2 + sinθ1 sinθ2) + i(sinθ1 cosθ2 - cosθ1 sinθ2)])

The real part of this expression is:

Re(z1,z2-bar) = r1r2(cosθ1 cosθ2 + sinθ1 sinθ2)

Using the identity cos(θ1 - θ2) = cosθ1 cosθ2 + sinθ1 sinθ2, we can write:

Re(z1,z2-bar) = r1r2 cos(θ1 - θ2)

Now we can substitute this expression back into the original equation and get:

r1r2 cos(θ1 - θ2) = r1r2

Dividing both sides by r1r2, we get:

cos(θ1 - θ2) = 1

This equation is true if and only if θ1 - θ2 = 2πn for some integer n. In other words, θ1 = θ2 + 2πn, where n is an integer.

Therefore, we have proved that Re(z1,z2-bar) = modulus of z1 * modulus of z2 if and only if arg z1 = arg z2 + 2πn, where n is an integer.