(1) Given two complex numbers z,-r1(cosa + 1 sin a) and z2-r2(cos θ2 + sin θ(2), prove the following formula for the division of complex numbers without using the Quotient Theorem stated in the text equal to the square of the modulus.

The formula for the division of complex numbers is:

z1/z2 = [(r1 cos a - r2 cos θ2) + i(r1 sin a - r2 sin θ2)] / [r2^2 + r1r2(cos(a - θ2) + i sin(a - θ2))]

To divide two complex numbers, we need to multiply the numerator and denominator by the complex conjugate of the denominator. That is, we multiply z1 by r2(cos θ2 - i sin θ2) and z2 by r2(cos θ2 - i sin θ2). This gives us:

z1/z2 = [(r1/r2)cos a - cos θ2 + i((r1/r2)sin a - sin θ2)] / [(r2 cos θ2 - i sin θ2)(cos a + i sin a)]

Next, we simplify the denominator using the identity cos^2θ + sin^2θ = 1:

z1/z2 = [(r1/r2)cos a - cos θ2 + i((r1/r2)sin a - sin θ2)] / [r2(cos a cos θ2 + sin a sin θ2) + i(r2 sin a cos θ2 - r2 cos a sin θ2)]

Now, we multiply the numerator and denominator by the conjugate of the denominator:

z1/z2 = [(r1/r2)cos a - cos θ2 + i((r1/r2)sin a - sin θ2)] / [r2(cos a cos θ2 + sin a sin θ2) + i(r2 sin a cos θ2 - r2 cos a sin θ2)] * [r2(cos a cos θ2 + sin a sin θ2) - i(r2 sin a cos θ2 - r2 cos a sin θ2)] / [r2(cos a cos θ2 + sin a sin θ2) - i(r2 sin a cos θ2 - r2 cos a sin θ2)]After simplifying, we get:

z1/z2 = [(r1 cos a - r2 cos θ2) + i(r1 sin a - r2 sin θ2)] / [r2^2 + r1r2(cos(a - θ2) + i sin(a - θ2))].

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We have proved the formula for the division of complex numbers without using the Quotient Theorem, which states that:

z1/z2 = |z1|/|z2| * (cos (θ1 - θ2) + i sin (θ1 - θ2))

To prove the formula for the division of complex numbers without using the Quotient Theorem, we can use the polar form of complex numbers and some trigonometric identities.

Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2) be two complex numbers in polar form.

Then, we can write the division of z1 by z2 as:

z1/z2 = r1(cos θ1 + i sin θ1) / r2(cos θ2 + i sin θ2)

Multiplying the numerator and denominator by the conjugate of z2, we get:

z1/z2 = r1(cos θ1 + i sin θ1) / r2(cos θ2 + i sin θ2) * r2(cos θ2 - i sin θ2) / r2(cos θ2 - i sin θ2)

Simplifying the numerator, we get:

z1/z2 = r1r2(cos θ1 cos θ2 + sin θ1 sin θ2 + i(sin θ1 cos θ2 - cos θ1 sin θ2))

Using the identities cos (θ1 - θ2) = cos θ1 cos θ2 + sin θ1 sin θ2 and sin (θ1 - θ2) = sin θ1 cos θ2 - cos θ1 sin θ2, we can write:

z1/z2 = r1r2(cos (θ1 - θ2) + i sin (θ1 - θ2))

Now, we can write z1/z2 in polar form as:

z1/z2 = |z1/z2| (cos φ + i sin φ)

where |z1/z2| = r1/r2 and φ = θ1 - θ2.

We can also see that the relation R does not have the comparability property, since for some complex numbers z1 and z2, it is not true that either z1 R z2 or z2 R z1. For example, if z1 = 1 + i and z2 = -1 - i, then z1 R z2 since |z1| < |z2|, but z2 does not R z1 since |z2| < |z1|. Therefore, R is not a total order on the set of complex numbers.

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