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Dividing complex numbers: polar & exponential form

Sal shows how complex division affects the modulus and the argument of the divisor and the dividend. Created by Sal Khan.

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  • piceratops ultimate style avatar for user Evan Indge
    At , Sal mentions "Euler's Formula", which is a foreign concept to me. Is there a video or videos that covers it, so I might be able to understand it a little better? Just the link should be fine... thanks!
    (18 votes)
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  • aqualine ultimate style avatar for user nobleman5055
    Can someone link me to a video were it shows how to use that variable e
    (11 votes)
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    • male robot hal style avatar for user Jesse
      Currently all you need to know about e is that it is a number. It is a really cool number, and it had some properties that are advantageous in calculus that make it the ideal base for an exponential or logarithmic function. But as far as manipulating exponentials with base e, you should use the same principles as you would with 2^x.
      (12 votes)
  • duskpin ultimate style avatar for user Natalie
    What is e? How would one define the constant?
    Also, can someone teach me more about Euler's formula? Isn't that a more advanced calculus topic?

    I am working on precalculus now and I found this topic. I'm kind of confused although everything else was crystal clear up until this point.
    (10 votes)
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    • purple pi purple style avatar for user APDahlen
      Hello Natalie,

      As an electrical engineer this is my favorite constant! It makes the math very easy. Wiki “phasor” if you are interested...

      Here is a joke for you - How many mathematicians does it take to change a light bulb? Answer : -e^(i*pi)

      Wiki has a good explanation of “e” at https://en.wikipedia.org/wiki/E_(mathematical_constant)

      Here is a good video especially if you like the Simpsons https://www.youtube.com/watch?v=Yi3bT-82O5s

      Know that the constant “e” alive and well. Take a look at your calculator you will see “e^x” and its cousin the natural logarithm “ln”. You will even find e on financial calculators.

      BTW the answer to my opening question is one since e^(i*pi) + 1 = 0.

      Sorry I couldn’t give a better explanation but I hope the links will get you to a good place.

      Regards,

      APD
      (12 votes)
  • leaf blue style avatar for user Cochran
    What is Euler's formula and how does it work? This video is acting like I already know what it is.
    (8 votes)
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  • mr pants teal style avatar for user Wrath Of Academy
    What does this actually mean? (The math is straightforward for converting to polar form and then dividing exponents and subtracting coefficients.) But what 2-dimensional use is this division? He divided a point by another point, and got some seemingly unrelated 3rd point. Is there some geometric relationship that gives meaning to all this?
    (5 votes)
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    • piceratops ultimate style avatar for user trek
      Engineers actually use operations on complex numbers to simplify mathematics of complicated problems. Dr Math presents the interesting real world problem of a snowplow. The more snow the plow is pushing, the slower the plow moves because it is pushing a heavier load, therefore the rate of growth of the snow heap slows over time. Instead of trying to track all of the variables, engineers use parametric equations (independent variable time) to calculate A(t), where A = amount of snow at time t.

      Dr Math also provided the example of expanding
      (cos (x) + i * sin(x))^3
      . Hairy expansion. Using Euler's formula, you can reduce a multi-step process (where you might make a careless error at any or all steps) to this:
       since e^(ix) = cos(x) + i *sin(x) and (x^a)^b) = x^(ab),
      e^(ix)^3 = e^i(3x)= cos(3x) + i * sin(3x)
      (5 votes)
  • female robot grace style avatar for user Achala
    as soon as sal sees the angle 7pi/6 he plots that angle . how?
    can any one tell me where i could find a video to teach that stuff
    (2 votes)
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    • piceratops ultimate style avatar for user Just Keith
      With radians, you get something easy happening, so that you know how much of a circle the angle sweeps out.
      number of radians = (fraction of a circle the angle sweeps out) (2π)
      So, just drop the π and divide by 2 and you get the fraction of the circle the angle sweeps out. (Of course, if there is no π in the angle, this isn't easy because you have to divide by π in that case.)

      And, once you know what fraction of a circle the angle sweeps out, it is quite easy to graph it.

      For example. How much of a circle does the angle 3π/2 sweep out?
      Dropping the π and dividing by 2 gives us ¾ of a circle.

      Another example: how much of a circle does the angle ⁵⁄₄ π sweep out? Its ⅝ of a circle.

      This is but one of many reasons why using radians is far easier than using degrees. With degrees, you have to divide by 360 to find out how much of a circle is swept out by an angle. And obviously finding half of something is far easier than dividing by 360.
      (6 votes)
  • piceratops ultimate style avatar for user Aidan Ji
    Couldn't You also write this polar form into rectangular form and then conjugate?
    (4 votes)
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  • orange juice squid orange style avatar for user Evan
    Could you just take the formula at , convert it to rectangular form (a + b * i), and then divide each of the parts?
    (2 votes)
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  • aqualine ultimate style avatar for user Ellie Jiang
    Is there is a specific pattern that emerges when you plot the point of the solution of dividing two complex numbers in relation to the original numbers? For example, at , Sal plots the solution on the imaginary plane, but is there a specific pattern that emerges as to the position of that point?
    (2 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      Good question! The answer is yes!

      The standard trigonometric angle (to the positive x-axis) associated with the quotient of two complex numbers, equals the difference of the standard trigonometric angles associated with them.

      The magnitude (or distance from the origin) of the quotient of two complex numbers is the quotient of their magnitudes.
      (2 votes)
  • blobby green style avatar for user ejcabanban
    In the powers of complex numbers question section their is an example question that asks:
    1) Find solution of following equation whose argument is between 225 deg - 315 deg ... z^7 = 128i

    Now when I click to get a hint it tells me that the modulus is 128 and that the argument is 90 + k * 360 . My question is how do you find that argument?
    (2 votes)
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    • aqualine ultimate style avatar for user David
      So we have z^7 = 128i.
      This is the same as z^7 = 128 (0 + 1*i).
      Let x be the argument.
      When is cos(x) = 0 and sin(x) = 1?
      (With help from the unit circle) this is when x = 90 + k * 360 degrees.
      Therefore, the argument is 90 + k * 360 degrees and we can write:
      z^7 = 128i = 128 (cos(90 + k * 360) + i * sin(90 + k * 360))
      and continue to solve the problem.
      (2 votes)

Video transcript

Voiceover:So this kind of hairy looking expression, we're just dividing one complex number, written in blue, by another complex number. This first complex - actually, both of them are written in polar form, and we also see them plotted over here. This first complex number, seven times, cosine of seven pi over six, plus i times sine of seven pi over six, we see that the angle, if we're thinking in polar form is seven pi over six, so if we start from the positive real axis, we're gonna go seven pi over six. So we're gonna go seven pi over six, all the way to that point right over there. And then from the origin, we're going to step seven out, or seven away from the origin. So one, two, three, four, five, six, seven to get to that point right over there. And then the second number, the angle is seven pi over four. So that takes us all the way around here. It takes us all the way - so I should do it smaller since that has a smaller distance, so let me - so let's say we start over here, so we're going to go all the way over here to that point right over here, and it's distance from the origin is one. You can imagine that there's a one in front of that. And we want to divide the two, and I encourage you to pause this video and try to do this on your own, and then plot the resulting number when you divide this blue complex number by this green one. Well, as you might have realized, if you somehow just try to divide it straight up, it can get quite hairy, and the way to tackle it, is to write it in another form. And what might have jumped out at you is that exponential form would be much, much simpler. And the way that we convert this to exponential form is to recognize that this business right over here is the same thing, this comes straight out of Euler's formula, this is e, to the seven pi over six i. Seven pi over six i. That's this expression right over here. So this entire top complex number can be rewritten as seven e, to the seven pi over six i. And this bottom complex number can be rewritten as one times, we didn't really have to write the one, but this bottom part right over here is going to be the same thing as e, to the seven pi over four i. Seven pi over four i, and this comes straight out of Euler's formula. And when you write it in this way, then we can just use exponent properties to simplify it. We have the same base, and so we can just subtract this exponent from that exponent right over there. So this is going to be equal to seven divided by one, is just seven. So it's going to be seven to the, or seven times e. Let me do this in a brighter color. Seven times e, to the seven pi over six i, minus seven pi over four i. Minus seven pi over four ith power. So what is this going to be equal to? Well if I have seven pi over six of something and I subtract seven pi over four of that thing, how many do I have left over? Well, let's see, I'm essentially - let's just rewrite all of this. This is really just about subtracting fractions at this point. So if I were to rewrite this, let's see, if I were to rewrite it with a denominator of 12, then I'll have a common denominator. It's my least common multiple of six and four. So this one I can rewrite as 14 pi i over 12. Or I can write it as 14 pi i over 12, and then minus - so I multiplied the numerator and denominator by three, minus 21 pi i over 12, as well. Right? I just multiplied numerators and denominators by three right over here. And so this is going to be equal to, I'm going to have, 12 in the denominator, 14 pi i, minus 21 pi i, is going to be negative seven pi i. So, this thing is equal to seven e, to the negative seven pi i, over 12. Now let's see if we can do a decent job of plotting this. So let's see, each of these increments that they've done, in each of these quadrants, we have one, two, three, four, five, six. So they split this quadrant into six equal angles, and this quadrant is pi over two, so each of these are pi over 12. Each of these little subangles are pi over 12. So we're going to have negative seven pi over 12. So we're going to go in the negative direction. We're going to go counter-clockwise, or sorry, we're going to go clockwise. So we're going to go, let me start here, so if I go one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, gets us right over there. Did I do that right? That doesn't - let me make sure I got that - negative seven, so each of these are pi over - let me make sure. So this whole thing is pi, so we have one - let me make sure, this whole thing is pi. We have one, two, three, four, five, six, seven, eight, nine, 10, 11, 12. So each of these are a twelfth of that, so it's pi over 12, that's right. Then we want to go negative seven of them. Negative one, negative two, negative three, negative four, negative five, negative six, negative seven, oh ya, I just kept going all the way to negative 12 the last time. So this is the angle, this is our angle right over here. And our distance that we go out from the origin is seven, so we go out one, two, three, four, five, six, seven, so we come out right over there. So this complex number divided by that complex number is equal to this complex number, seven times e, to the negative seven pi i over 12. And if we wanted to now write this in polar form, we of course could. We could say that this is the same thing as seven, times cosine of negative seven pi over 12, plus i sine of negative seven pi over 12.