Hello everyone. In this equation, I'm going to do some complex numbers.
So I should warn you that I didn't actually double check my answer, but I'm
pretty sure it's correct. I'll leave it up to you guys to double check my work, and if I'm wrong
let me know, and I'll reproduce the video.
Determine all complex numbers, in polar or standard form, of the equation
z to the power of 6 plus 2i z cubed minus 4 equals 0.
So I'm going to do this in
probably polar form by the end,
but I'll probably start in standard form. So
So when I look at this equation, I do see
immediately a quadratic, right? z to the 6 plus something times z cubed minus 4.
This is basically a quadratic if I replace z to the cubed with a w,
so I'm going to that. So let w equal z cubed so that
w squared plus 2i w minus 4 equals 0.
By the way, it should be noted that you should try to factor this first,
and I tried to factor it and I didn't find the factors by hand.
So when you don't find the factors by hand, you can use the quadratic formula to get them.
So let's plug it into the quadratic formula.
So w is equal to negative b plus or minus the square root of b squared minus 4 a c over 2a.
Plug it in, simplify, and we get something pretty nice. You get
minus i plus or minus the square root of 3.
Now with this, I'm going to take this and I'm going to try
to solve the final equation. So now I have w is equal to this,
and w is z cubed, so now it suffices to solve z cubed is equal to minus i
minus the square root of 3.
So we should right now immediately be thinking oh I should be using CNRT.
I have a cubic thing is equal to some complex number,
so I should be able to take cube roots by converting to polar form and transferring, okay?
So I'm going to take our two equations
and I'm going to do the switcheroo that I did in class, right?
So the way I like to convert these is
by taking out the modulus, right, because I need to write this as r
times something with norm 1, okay?
So compute the modulus, so compute the length of these two vectors.
It's going to turn out to be 2. It doesn't matter if I use
plus or minus the square root of 3 because I square the term inside the square root.
So plus or minus the square root of 3 squared is going to be the same
Maybe I'll actually even put that in just so that it's a little bit clearer.
So there we go.
So it's going to give me 1 plus 3 which is
4, square root of 4 is 2,
and thus we're solving these two equations given by z cubed is equal to 2,
so I'm going to take out the modulus part,
plus or minus root 3 over 2 minus i over 2.
Now this should look like some sort of nice trig identity, and in fact it is.
So again, there's no shame, draw the little picture, and figure out what angle this should be.
So if I do the negative cosine - so negative square root of 3 over 2 minus i over 2.
So the cosine and sine are negatives that happens in the third quadrant.
Cosine is root 3 over 2, so my adjacent should be root 3,
and my hypotenuse should be 2. So that gives me this triangle,
and that angle is 7 pi over 6. Remember shortest angle goes with the shortest side,
and similarly for positive square root of 3 over 2.
That's going to be - so the cosine is positive
and the sine is negative, that's quadrant 4.
Again because that angle over here is
pi over 6, yeah I guess my diagram is not really labeled that well.
This should be 11 pi over 6 on the other way.
So let me…
let me fix that. Just give me a quick second to fix that little picture.
Okay, so let's save that and let's reload the picture.
So let me make sure I get my angles correct, like so.
So the big angle is going to be 7 pi over 6, it's going to be
pi plus pi over 6,
and the last one's going to be 2 pi minus the pi over 6, so 11 pi over 6.
So now we have the two equations given by z cubed is equal to 2 times
cis of 7 pi over 6 or
cosine 7 pi over 6 plus i sine 7 pi over 6,
and z cubed is equal to 2 times cosine of 11 pi over 6 plus i sine 11 pi over 6.
So some point of confusion is a lot of people think
oh well I had a negatives up here and where did the negatives go?
Remember that sine of 11 pi over 6 is negative,
and sine of 7 pi over 6 is negative,
and cosine of 7 pi over 6 is negative. So the signs are taken care of by the trig functions.
So thus, now all I have to do is just plug in CNRT.
So what does CNRT say? Okay it says take the
cube root of the modulus, and then divide these two angles by 3,
and then add multiples of 2 pi,
and that gives you all of them. Now you could simplify this a little bit
I chose not to just to show the form,
but again this is just a quick application now of CNRT.
So once you get it into this nice form, CNRT can take you home.
So hopefully that gave you a little bit of an idea of what's going on.
So again make sure you understand how to use CNRT.
So once you write it in polar form, divide the angles by the degree,
and take the nth root of the r value,
and just write down your solutions. Now the question said I can put it in polar or standard form,
so I'm going to put it in polar form.
I think that’s perfectly reasonable, and
it seemed to fall out naturally for me from that form.
Okay great. So thank you very much for your time.
Hopefully this video was a little bit informative, helped you review some complex numbers.