^{iθ}

^{2}+ y

^{2}) and θ = tan

^{-1}(y/x)

^{iθ}) = (1/r) e

^{-iθ}

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Yet another way:

You can express a complex number z = x + iy in polar form as

z = re^{iθ}

where r = √(x^{2} + y^{2}) and θ = tan^{-1}(y/x)

Then dividing by z is the same as multiply by 1/z, where

1/z = 1/(re^{iθ}) = (1/r) e^{-iθ}

This example brings me back to the first run through DeMoivre's Theorem. A great way to introduce Inductive Proofs! A recommendation for building good creative mathematical skills. I tend toward more familiar trigonometric forms with Moduli and Amplitude excluding usage of "e." All forms are valid, nice post!

The standard method is to multiply the numerator & denominator by the conjugate of the complex denominator.

The denominator is then always a real number.

Hi John,

I am guessing the intent of this problem was to describe how to divide (6 - 3i) by (2 + 7i), for example.

That would be (6 - 3i)/(2 + 7i). The process is to multiply numerator and denominator of the fraction by the conjugate of the denominator (2 - 7i). This works because the product in the denominator results in a difference of squares. Since i^{2} = -1, the imaginary part of the complex number in the new denominator disappears. The result is a new complex number (the quotient).

I guess the property that justifies the process is that a number/expresssion multiplied by the multiplicative identity (1) results in a number/expression with the same value. And (2 - 7i)/(2-7i) does equal one (multiplicative identity).

So:

(6 - 3i)(2 - 7i)/(2 + 7i)(2 - 7i) = ? I'll let you take it from there.

The Complex numbers form what is called a Field in the theory of groups.

Satisfying the 11 properties of a Field, albeit not being an ordered field in that the square power of the imaginary unit is -1. They do not hold the trichotomy property.

This will result that imaginaries are not considered greater than or less than other elements of the universe set of pure imaginary or complex numbers. They are considered only as Vectors with Magnitudes.

Because Closure, Assossiativity and Commutivity under two Operations; Addition and Multiplication namely, and Existence of Identity Elements for Addition and Multiplication along with inverse elements for these respective Operations, however excluding 0 + 0i, and additionally that Complex Numbers are Distributive with respect to Multiplication over Addition, we can define in algebraic terms that every operation on dividing by a complex number will yield a complex number belonging to the Universal Set of Complex Numbers.

This is why the method works in the sense you will always produce a real solution or complex solution to a division problem with complex numbers in the denominator. The algebra forces a functional output belonging to the Complex Numbers in every infinitely possible case.

It is a complicated way of saying, "When adding apples and oranges, (or multiplying them), you are always left with some countable discreet number of apples and oranges, no bananas allowed!" (This is why they do not include division by zero.)

Peter P.

If you ever divided a polynomial by a monomial, the you know you divide each term in the polynomial by that monomial and add up those terms. In this case, the complex number, a+bi, is the polynomial.

Take the solution of a quadratic equation using the quadratic formula for example.

x = (2 ± 6i) / 2

x = 1 ± 3i

The method used here is distributive property.

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