Quotient of Two Complex Numbers Given two complex numbers z1=r1(cosθ1+i sinθ1) and z2=r2(cosθ2+i

Quotient of Two Complex Numbers Given two complex numbers...

Key Concepts

Division of Complex Numbers in Polar Form

Dividing complex numbers in polar form involves two key operations: dividing the moduli and subtracting the arguments. This means if you have two numbers represented as r?(cos ?? + i sin ??) and r?(cos ?? + i sin ??), their quotient is given by (r?/r?)[cos(?? - ??) + i sin(?? - ??)], which simplifies the computation and illustrates the interplay between magnitude and angle in complex arithmetic.

Modulus and Argument

The modulus (or absolute value) of a complex number is the distance from the origin to the point representing the number in the complex plane, while the argument is the angle that line makes with the positive real axis. These concepts are crucial as they bridge the algebraic and geometric interpretations of complex numbers, providing deeper insights into their properties and operations.

Complex Numbers

Complex numbers, which take the form a + bi (where a and b are real numbers and i is the imaginary unit with i² = -1), form the basis of much of mathematics dealing with two-dimensional number spaces. They enable solutions to equations that have no real solutions and serve as a foundational element in fields such as engineering, physics, and applied mathematics.

Polar Form Representation

The polar form represents a complex number in terms of its modulus and argument, expressed as r(cos ? + i sin ?). This representation emphasizes the geometric interpretation of complex numbers as points in the plane, where r is the distance from the origin and ? is the angle measured from the positive real axis, making calculations involving multiplication and division more intuitive.

Transcript

00:01 For this problem, we are given that z1 equals r1 cost theta 1 plus i sine theta 1, and z2 equals r2 cos theta2 plus i sign theta 2, we want to prove that z1 over z2 times cost theta 1 minus theta 2 plus i sine theta 1 minus theta 2.

00:19 So the way that we start out for this, we can write this out just explicitly based on what we are given up above.

00:28 In fact, i will just copy and paste down what i have.

00:31 Up above that z1 equals r1 cost theta 1 plus i sine theta 1 and then we are dividing by r2 cost theta 2 plus i sine theta 2 now what we want to do here is we can multiply the top and the bottom by the complex conjugate of the denominator so out front we'll keep that r1 over r2 that's easy enough and then on top we'll have actually, i'll do a little bit more copying and pasting to save my wrist here.

01:14 I sign theta 1 times.

01:22 Now i'll copy and paste what i had in the denominator, but i need to change that plus over to a minus since we are multiplying by the complex conjugate.

01:32 But keep in mind that since we are multiplying both the numerator and denominator of this fraction by the same thing, the complex conjugate of the denominator, we're not actually changing anything.

01:43 We're doing the same thing as multiplying by 1.

01:46 It's just that we're writing 1 in a very peculiar way.

01:51 Any number divided by itself is going to equal 1, even if it is the complex conjugate of the denominator.

01:59 All right.

02:01 And now that i don't need to find ways to keep talking, we have this written out.

02:06 Now, now that we have that done, you can go about multiplying together the numerator and denominator.

02:15 So on top, you'll get costeatimilar.

02:17 Times cost theta 2 minus i sine theta 2 times cost theta 1 which i'll write as minus i coast theta 1 times sine theta 2 then we'll have plus i sine theta 1 times coast theta 2 and then we'll have minus i squared sine theta 1 sine theta 2 then i know unfortunately there's not enough space on screen to fit this whole thing in here.

03:05 That's a little bit too thick.

03:09 Okay.

03:10 Then, on the denominator, we'll get cost squared theta 2 minus cost theta 2, or minus i, rather, i coast theta 2 times sine theta 2, plus i coast theta 2, sine theta 2, plus i coast theta 2, sine squared theta 2.

03:49 Now we can immediately see that those middle two terms in the denominator are going to add up to zero.

03:56 And we can do some rearranging in the numerator as well.

04:00 We still have that r1 over r2 out front unchanged...

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