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Hello everyone. In this equation, I'm
going to do some complex numbers.

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So I should warn you that I didn't actually 
double check my answer, but I'm

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pretty sure it's correct. I'll leave it up to you 
guys to double check my work, and if I'm wrong

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let me know, and I'll reproduce the video.

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Determine all complex numbers, in 
polar or standard form, of the equation

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z to the power of 6 plus 2i 
z cubed minus 4 equals 0.

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So I'm going to do this in

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probably polar form by the end,

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but I'll probably start in standard form. So

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So when I look at this equation, I do see

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immediately a quadratic, right? z to the 6 
plus something times z cubed minus 4.

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This is basically a quadratic if I 
replace z to the cubed with a w,

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so I'm going to that. So let 
w equal z cubed so that

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w squared plus 2i w minus 4 equals 0.

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By the way, it should be noted that 
you should try to factor this first,

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and I tried to factor it and I 
didn't find the factors by hand.

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So when you don't find the factors by hand, 
you can use the quadratic formula to get them.

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So let's plug it into the quadratic formula.

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So w is equal to negative b plus or minus the
square root of b squared minus 4 a c over 2a.

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Plug it in, simplify, and we get 
something pretty nice. You get 

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minus i plus or minus the square root of 3.

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Now with this, I'm going to 
take this and I'm going to try

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to solve the final equation. So 
now I have w is equal to this,

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and w is z cubed, so now it suffices 
to solve z cubed is equal to minus i

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minus the square root of 3.

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So we should right now immediately be 
thinking oh I should be using CNRT.

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I have a cubic thing is equal 
to some complex number,

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so I should be able to take cube roots by 
converting to polar form and transferring, okay?

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So I'm going to take our two equations

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and I'm going to do the switcheroo 
that I did in class, right?

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So the way I like to convert these is

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by taking out the modulus, right, 
because I need to write this as r

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times something with norm 1, okay?

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So compute the modulus, so compute 
the length of these two vectors.

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It's going to turn out to be 
2. It doesn't matter if I use

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plus or minus the square root of 3 because 
I square the term inside the square root.

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So plus or minus the square root of 
3 squared is going to be the same

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Maybe I'll actually even put that in 
just so that it's a little bit clearer.

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So there we go.

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So it's going to give me 1 plus 3 which is

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4, square root of 4 is 2, 

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and thus we're solving these two 
equations given by z cubed is equal to 2,

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so I'm going to take
out the modulus part,

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plus or minus root 3 
over 2 minus i over 2.

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Now this should look like some sort 
of nice trig identity, and in fact it is.

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So again, there's no shame, draw the little 
picture, and figure out what angle this should be.

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So if I do the negative cosine - so negative 
square root of 3 over 2 minus i over 2.

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So the cosine and sine are negatives 
that happens in the third quadrant.

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Cosine is root 3 over 2, so my
adjacent should be root 3,

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and my hypotenuse should be 2. 
So that gives me this triangle,

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and that angle is 7 pi over 6. Remember 
shortest angle goes with the shortest side,

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and similarly for positive
square root of 3 over 2.

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That's going to be - so the cosine is positive

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and the sine is negative, that's quadrant 4.

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Again because that angle over here is

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pi over 6, yeah I guess my diagram 
is not really labeled that well.

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This should be 11 pi
over 6 on the other way.

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So let me…

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let me fix that. Just give me a 
quick second to fix that little picture.

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Okay, so let's save that and
let's reload the picture.

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So let me make sure I get 
my angles correct, like so.

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So the big angle is going to be 
7 pi over 6, it's going to be

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pi plus pi over 6,

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and the last one's going to be 2 pi 
minus the pi over 6, so 11 pi over 6.

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So now we have the two equations 
given by z cubed is equal to 2 times

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cis of 7 pi over 6 or

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cosine 7 pi over 6 plus i sine 7 pi over 6,

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and z cubed is equal to 2 times cosine 
of 11 pi over 6 plus i sine 11 pi over 6.

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So some point of confusion 
is a lot of people think

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oh well I had a negatives up here 
and where did the negatives go?

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Remember that sine of 
11 pi over 6 is negative,

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and sine of 7 pi over 6 is negative,

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and cosine of 7 pi over 6 is negative. So the 
signs are taken care of by the trig functions.

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So thus,  now all I have to do is just plug in CNRT.

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So what does CNRT say? 
Okay it says take the

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cube root of the modulus, and
then divide these two angles by 3,

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and then add multiples of 2 pi,

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and that gives you all of them. Now 
you could simplify this a little bit

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I chose not to just to show the form,

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but again this is just a quick 
application now of CNRT.

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So once you get it into this nice 
form, CNRT can take you home.

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So hopefully that gave you a little 
bit of an idea of what's going on.

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So again make sure you
understand how to use CNRT.

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So once you write it in polar form, 
divide the angles by the degree,

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and take the nth root of the r value,

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and just write down your solutions. Now the 
question said I can put it in polar or standard form,

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so I'm going to put it in polar form.

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I think that’s perfectly reasonable, and

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it seemed to fall out naturally 
for me from that form.

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Okay great. So thank you 
very much for your time.

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Hopefully this video was a little bit informative, 
helped you review some complex numbers.