Simplify \(\sqrt{7+24i}\)

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Simplify \(\sqrt{7+24i}\)

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- dan815

how about just
Re^itheta = 7+24i

- Empty

I don't know

- dan815

hm thats not nice

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## More answers

- dan815

|dw:1434880616850:dw|

- dan815

the angle seems to be exact in degrees

- Empty

Ok wait I figured it out, but it doesn't seem natural:
\[\sqrt{7+24i}=\sqrt{(a+bi)^2}\]
\[7+24i = a^2-b^2 + 2abi\]
now we split this up into two equations with two unknowns
\[7=(a+b)(a-b)\]\[12=ab\] since 7 is positive and prime that means a-b=1 and a+b=7 so we have:
a=4, b=3
\[\sqrt{7+24i} = 4+3i\]
Although I guess it didn't really have to work out this way, they could have been rational numbers for all I know or something.

- dan815

oh thats a nice trick

- Empty

Next one looks like \[\sqrt{-8-6i}\] I did the same thing basically but it looks like I'm getting two answers.

- Empty

I'm getting that it's equal to both \[1-3i\] and \[-1+3i\]

- dan815

there could be 2 answers

- dan815

going forward and backwards

- dan815

|dw:1434881253505:dw|

- dan815

both of them look about right

- dan815

i wonder if there is another way to do this, like how about normalizing and finding the half angle first

- Empty

Makes sense since one is the negative of the other, we could just factor out like this:
\[\sqrt{-8-6i}=\pm i\sqrt{8+6i}\]

- dan815

and then sqrting the radius

- dan815

then convert that to cartesian

- dan815

or would that get messy with e and pi in that mix

- Empty

I think it might, unless you are able to recognize a pythagorean triple's angle which might not be immediately apparent I think. Not sure

- Empty

I think this ties in nicely with what you were saying earlier, what does the square root of this complex number mean...?

- dan815

for some reason i always see this anon stalker in my questions all the time http://prntscr.com/7ji57g

- Empty

Is there a visual way in which we can consider the square root of -8-6i as being 1-3i or -1+3i.

- dan815

well ya, like half way from origin and position x axis , and if i see the magnitude of the radius double upto the main radius

- Empty

I guess that's what you did, you drew it then rotated in either direction

- dan815

yeah

- Empty

and then took the square root of the length

- dan815

right ya squaring i mean not doubling, i confuse those 2

- Empty

Yeah I gotcha, ok here's a good one looks kinda simple though http://prntscr.com/7ji5ww

- dan815

we could see they are all orthogonal or

- dan815

wait we can do that thing from before, normalize all and all the vectors can be written in that e^i2pik/4 form

- dan815

or just take dot products and that bs

- Empty

We could subtract one of the vectors from the other 3 to make that point the origin with 3 points coming off it

- Empty

then we know that one of the vectors must be 90 degrees of the other, so we kinda maybe figure that out and rotate

- dan815

is there a way to rewrite all these points in a rotated dimension, like a simple way

- dan815

like if i wanna just rewqrite all these vectors but with the x and y axis rorated to one of the vectors

- Empty

Once we subtract one from the rest, we just plug in the two furthest ones apart into matrix and take the determinant, square root that and that's the side length, multiply that by sqrt(2) and that should be the distance to the last vector right?

- dan815

oh my bad i think i am confusing this question, for some reason i was thinking they are giving us vectors..

- dan815

they just gave us points ah

- Empty

Points, vectors, whatever

- dan815

yea this still works i guess

- dan815

- dan815

that will turn them into vectors so we can do this agian

- dan815

the problem before was it cud have been like a square placed weirdly on the axis, so just dotitng their points wudnt make any sense

- dan815

|dw:1434882161951:dw|

- Empty

Woah once you shift it by subtracting 3i from all the points you get the three vectors:
2-2i, 4, 2+2i which is |dw:1434882136744:dw|

- Empty

|dw:1434882255217:dw| Since those are the midpoints on a square, this is a square.

- Empty

Their solution is ugly as crap: http://www.exampleproblems.com/wiki/index.php/CV19

- dan815

|dw:1434882335886:dw|

- Empty

No danny don't break space and do that infinitely otherwise we'll prove that points are really squares

- dan815

oh thats gross they are writing it out as lines lol

- dan815

haha

- Empty

Yeah they're doing it the way I would program a computer to do it maybe, but I have a brain tyvm

- Empty

maybe if they had presented it as arbitrary points or something, actually maybe we should solve the arbitrary case. I think we could do it pretty easily.

- IrishBoy123

why aren't you using complex exponentials
first one: \(z = 25 e^{i \theta}, tan \theta = 24/7\)
\(\sqrt{z} = \pm 5 e^{i \theta/2}\)

- dan815

i think a faster way if u gonna program this stuff would bne
if its labelled
properly
p1 p2 p3 p4
find the 4 mags
p1-p2, p2-p3,p3-p4,p4-p1
then
p1-p2 dot p2-p3, p2-p3 dot p3-p4....
if all the mages are equal and all dots are 0 then done

- dan815

this way u can break out of the loop as soon as one if it fails

- Empty

Here's one way to write out the corners of a square with corners at a,b,c,d:
\[a=re^{i \theta}+z\]\[b=rie^{i \theta}+z\]\[c=-re^{i \theta}+z\]\[d=-rie^{i \theta}+z\]
r is the distance from the center to a corner of the square. \(\theta\) is the amount of rotation we want to give it, so it really only matters to vary it from \(0 < \theta < \pi /2\) since it'll just repeat itself and z is a complex number that translates it.

- Empty

So 4 equations with 4 unknowns, r, \(\theta\) are just real numbers and z has 2 numbers since it's complex. Coolio.

- dan815

yeah, that wud be nice, but we wont always get lucky

- Empty

What do you mean why not?

- dan815

they can give us annoying points that arent centered

- Empty

They don't have to be centered for this to work

- dan815

find we have to take the double diagonal and reposition that intersection as the origin

- Empty

Woah! I just saw something miraculous

- Empty

If it's a square then we can add up all the 4 complex numbers to get the center of the square.

- dan815

wait really

- Empty

Then we subtract this from the 4 numbers to center it.

- dan815

no matter where the squre is placed?

- Empty

It's like vectors in equilibrium.

- Empty

But the equilibrium is the center displaced

- Empty

Let's try it on the example they have
2+i, 4+3i, 2+5i, 3i is a square. Add them all together and subract the sum from all of them.

- dan815

|dw:1434883019314:dw|

- Empty

Then we can literally square any of the numbers we have to get the other 3, after we normalize them.

- Empty

|dw:1434883081383:dw|

- dan815

yaa
thats what i mean though, u wont get nice vertices though

- Empty

watch, I'll do it.

- dan815

ohh wait! ur saying additiong of these 4 vectors will give u center, i can see it kind of :O

- dan815

woah thats cool, it makes sense it has to go toward the center

- Empty

Yeah this is the best way for a computer to do it too for any polygon.

- dan815

the farther away from center u are the more u are pulled to the otehr vertices

- Empty

Add up all the vertices, subtract this number from all of them, normalize them, and take one arbitrary one and square it until you get all the way around the polygon. If you don't hit one then you you just pick another one cause you happened to pick a root of 1 that wasn't relatively prime or whatever.

- dan815

this gives us a much simpler way of solving the square question now

- dan815

this works for all regular polygons

- Empty

Yeah

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Because all regular n-gons are just the n-roots of 1 stretched out by some radius and then translated in the plane somewhere from this perspective essentially.

- Empty

Wait this doesn't quite work hahaha

- Empty

It should be the average of the points not just the sum of the points.

- dan815

ya i was just looking at that lol

- dan815

|dw:1434883586373:dw|

- Empty

\[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\]
Subtract that from each of those same 4 points:
-2i, 2, 2i, -2.

- Empty

What could be sexier?

- dan815

|dw:1434883777297:dw|

- dan815

like thats an easy one to see that fails for sure

- Empty

No, that doesn't fail

- dan815

why not

- Empty

\[\frac{0+2-2i+(2-2i)}{4}=1+i\]

- dan815

oh ya i mean not this, the one wiht just adding

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Oh yeah, we're over that now, the thing that makes it clear is just imagine finding the center point of each dimension independently, that's just the average of each independently so since they don't mix when they add their averages stay separate which is nice.

- dan815

|dw:1434883967181:dw|

- dan815

this is the same thing just said a bit differently

- dan815

with the 4 vectors 2 at a time share a b c d

- Empty

True but your method fails if the square looks like this: |dw:1434884123357:dw|

- dan815

oh truee

- dan815

well its something thoughj still as long as the point is somethere in the square

- Empty

Actually specifically for the square I see something special we can do after we move a square like this to the center. We can instantly check by just comparing that the real and imaginary parts are all identical but with a difference of + or - or i.

- Empty

|dw:1434884274366:dw|

- dan815

lol

- Empty

No offense....

- Empty

lolol

- dan815

we can also just shift the square by some point, so that the origin is located anywhere in the square

- Empty

But you can see that there will be uniquely only two numbers for each of the four vectors after translating it to the center.

- Empty

Yeah but dan do you not understand what I said before?

- dan815

which one?

- Empty

Like once we average the points no matter where it was, that will give us the center. Once we subtract this value out, it doesn't matter how the square is rotated.

- dan815

oh ya i can see that one works no matter what

- dan815

im just seeing if there is a less computation way in some cases

- dan815

not that it matters lol

- Empty

I think this is the least computational way unless you wanna do it their gross way lol

- dan815

well like (-a+b, -c+d) is center is slightly faster than (a,c) + (a,d) + (b,d)+ (b,c)/ 4

- dan815

only works if origin is somewhere in square.. it doesnt make a difference, but i can see how maybe in computer programs.. it might add up eventually if someone knows all their calculations involve origin in square

- Empty

I don't understand how your calculation is realistic though

- dan815

why not

- Empty

I'm not saying this calculation is what I'm doing, you have a misconception if you think that (a,c) + (a,d) + (b,d)+ (b,c)/ 4 is what my calculation does. There is no relationship like this, in general there is not a repetition that occurs like this, just look at the example we got our question from it was this which happened to only have repetition on only 2 entries and even that was because the square was rotated perfectly 45 degrees i believe.\[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\]

- Empty

In that example what is a,b,c, and d if there are coefficients 2,1,4,3,5,0 ?

- dan815

ohh i see the problem okayy

- dan815

ah okay that makes sense

- dan815

the rotations

- dan815

ok ya i see, dont know why i was thinkign the complex numbers would be repeating we can have all unique points and that summing and dividing by 4 still works

- dan815

they should allow rotation in the draw tool, that give so many more options

- Empty

If the number of sides on the polygon is a multiple of 4 and they're regularly spaced then we can see that after we subtract this center vector we will have:
\[k=2*\lceil \frac{(number \ of \ vertices)}{8} \rceil\]
Where k is the number of unique vector components we'll have... See if you can figure this one out ;)

- dan815

what do you mean reguarly spaced and subtract the center vector

- dan815

|dw:1434885527859:dw|

- dan815

that vector?

- Empty

Imagine I give you the complex numbers representing a regular octogon that has been randomly thrown on the xy plane, rotated and translated anywhere. How many unique vector components (well real and imaginary parts) will it have when you shift it to the center by yes that vector you just drew, which is the average of the vertex vectors.

- dan815

ah okay that makes sense

- dan815

once u subtract the center vector every point is share by 2 vectors

- dan815

every componnent of the point

- dan815

wait no

- Empty

When it's a regular octogon centered at the origin but arbitrarily rotated then imagine it's really the vertices of two squares, one of which is the reflection of the other... lol

- dan815

are u sure this formula works

- dan815

a square at the center unrotated vs rotated would have tme same number of unique vector components with that formula, but it shouldnt

- dan815

does it have to more a multiple of 4 more than times 1

- dan815

have to be*

- dan815

|dw:1434886153355:dw|

- dan815

like for instance this one, where its upright perfectly will share 2 at a time, but if its rorated slightly they would all be unique

- dan815

or actually 4 at a time

- Empty

No, rethink this, you're on the right track though.

- dan815

|dw:1434886388250:dw|

- Empty

Ahhh ok yeah you're right it is only every 4 I was thinking every 8 I see

- dan815

i was cutting green peppers.. and i wiped my eyes

- dan815

ok ya i see now

- Empty

rofl

- dan815

i can see how u ll start start to have rotated vector components

- dan815

of pi/2 and -pi/2 and showing up

- dan815

so that means
a,b --- > -b,a

- dan815

and
a,b --- > b,-a

- dan815

for multiples of 4 the 90 degree ones must exist for every point

- Empty

Yeah I think this is the simplest way to check by a computer, although let's pretend you had a million sided regular polygon in the complex plane how would you check it

- Empty

ok nvm that's stupid why did I ask that lol

- Empty

http://prntscr.com/7jiujx
Let's do this one

- dan815

hey if u draw the tangents to 8 polygon u get 2 squares?

- dan815

|dw:1434886974869:dw|