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Empty
 one year ago
Simplify \(\sqrt{7+24i}\)
Empty
 one year ago
Simplify \(\sqrt{7+24i}\)

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dan815
 one year ago
Best ResponseYou've already chosen the best response.2how about just Re^itheta = 7+24i

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the angle seems to be exact in degrees

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ok wait I figured it out, but it doesn't seem natural: \[\sqrt{7+24i}=\sqrt{(a+bi)^2}\] \[7+24i = a^2b^2 + 2abi\] now we split this up into two equations with two unknowns \[7=(a+b)(ab)\]\[12=ab\] since 7 is positive and prime that means ab=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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Next one looks like \[\sqrt{86i}\] I did the same thing basically but it looks like I'm getting two answers.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I'm getting that it's equal to both \[13i\] and \[1+3i\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2there could be 2 answers

dan815
 one year ago
Best ResponseYou've already chosen the best response.2going forward and backwards

dan815
 one year ago
Best ResponseYou've already chosen the best response.2both of them look about right

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i wonder if there is another way to do this, like how about normalizing and finding the half angle first

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Makes sense since one is the negative of the other, we could just factor out like this: \[\sqrt{86i}=\pm i\sqrt{8+6i}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2and then sqrting the radius

dan815
 one year ago
Best ResponseYou've already chosen the best response.2then convert that to cartesian

dan815
 one year ago
Best ResponseYou've already chosen the best response.2or would that get messy with e and pi in that mix

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1I think this ties in nicely with what you were saying earlier, what does the square root of this complex number mean...?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for some reason i always see this anon stalker in my questions all the time http://prntscr.com/7ji57g

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Is there a visual way in which we can consider the square root of 86i as being 13i or 1+3i.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2well 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
 one year ago
Best ResponseYou've already chosen the best response.1I guess that's what you did, you drew it then rotated in either direction

Empty
 one year ago
Best ResponseYou've already chosen the best response.1and then took the square root of the length

dan815
 one year ago
Best ResponseYou've already chosen the best response.2right ya squaring i mean not doubling, i confuse those 2

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah I gotcha, ok here's a good one looks kinda simple though http://prntscr.com/7ji5ww

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we could see they are all orthogonal or

dan815
 one year ago
Best ResponseYou've already chosen the best response.2wait we can do that thing from before, normalize all and all the vectors can be written in that e^i2pik/4 form

dan815
 one year ago
Best ResponseYou've already chosen the best response.2or just take dot products and that bs

Empty
 one year ago
Best ResponseYou've already chosen the best response.1We could subtract one of the vectors from the other 3 to make that point the origin with 3 points coming off it

Empty
 one year ago
Best ResponseYou've already chosen the best response.1then we know that one of the vectors must be 90 degrees of the other, so we kinda maybe figure that out and rotate

dan815
 one year ago
Best ResponseYou've already chosen the best response.2is there a way to rewrite all these points in a rotated dimension, like a simple way

dan815
 one year ago
Best ResponseYou've already chosen the best response.2like if i wanna just rewqrite all these vectors but with the x and y axis rorated to one of the vectors

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Once 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
 one year ago
Best ResponseYou've already chosen the best response.2oh my bad i think i am confusing this question, for some reason i was thinking they are giving us vectors..

dan815
 one year ago
Best ResponseYou've already chosen the best response.2they just gave us points ah

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Points, vectors, whatever

dan815
 one year ago
Best ResponseYou've already chosen the best response.2yea this still works i guess

dan815
 one year ago
Best ResponseYou've already chosen the best response.2that will turn them into vectors so we can do this agian

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the problem before was it cud have been like a square placed weirdly on the axis, so just dotitng their points wudnt make any sense

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Woah once you shift it by subtracting 3i from all the points you get the three vectors: 22i, 4, 2+2i which is dw:1434882136744:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434882255217:dw Since those are the midpoints on a square, this is a square.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Their solution is ugly as crap: http://www.exampleproblems.com/wiki/index.php/CV19

Empty
 one year ago
Best ResponseYou've already chosen the best response.1No danny don't break space and do that infinitely otherwise we'll prove that points are really squares

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh thats gross they are writing it out as lines lol

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah they're doing it the way I would program a computer to do it maybe, but I have a brain tyvm

Empty
 one year ago
Best ResponseYou've already chosen the best response.1maybe 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
 one year ago
Best ResponseYou've already chosen the best response.0why 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
 one year ago
Best ResponseYou've already chosen the best response.2i think a faster way if u gonna program this stuff would bne if its labelled properly p1 p2 p3 p4 find the 4 mags p1p2, p2p3,p3p4,p4p1 then p1p2 dot p2p3, p2p3 dot p3p4.... if all the mages are equal and all dots are 0 then done

dan815
 one year ago
Best ResponseYou've already chosen the best response.2this way u can break out of the loop as soon as one if it fails

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Here'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
 one year ago
Best ResponseYou've already chosen the best response.1So 4 equations with 4 unknowns, r, \(\theta\) are just real numbers and z has 2 numbers since it's complex. Coolio.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2yeah, that wud be nice, but we wont always get lucky

Empty
 one year ago
Best ResponseYou've already chosen the best response.1What do you mean why not?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2they can give us annoying points that arent centered

Empty
 one year ago
Best ResponseYou've already chosen the best response.1They don't have to be centered for this to work

dan815
 one year ago
Best ResponseYou've already chosen the best response.2find we have to take the double diagonal and reposition that intersection as the origin

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Woah! I just saw something miraculous

Empty
 one year ago
Best ResponseYou've already chosen the best response.1If it's a square then we can add up all the 4 complex numbers to get the center of the square.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Then we subtract this from the 4 numbers to center it.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2no matter where the squre is placed?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1It's like vectors in equilibrium.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1But the equilibrium is the center displaced

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Let'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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Then we can literally square any of the numbers we have to get the other 3, after we normalize them.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2yaa thats what i mean though, u wont get nice vertices though

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ohh wait! ur saying additiong of these 4 vectors will give u center, i can see it kind of :O

dan815
 one year ago
Best ResponseYou've already chosen the best response.2woah thats cool, it makes sense it has to go toward the center

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah this is the best way for a computer to do it too for any polygon.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the farther away from center u are the more u are pulled to the otehr vertices

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Add 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
 one year ago
Best ResponseYou've already chosen the best response.2this gives us a much simpler way of solving the square question now

dan815
 one year ago
Best ResponseYou've already chosen the best response.2this works for all regular polygons

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Because all regular ngons are just the nroots of 1 stretched out by some radius and then translated in the plane somewhere from this perspective essentially.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Wait this doesn't quite work hahaha

Empty
 one year ago
Best ResponseYou've already chosen the best response.1It should be the average of the points not just the sum of the points.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ya i was just looking at that lol

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\] Subtract that from each of those same 4 points: 2i, 2, 2i, 2.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2like thats an easy one to see that fails for sure

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{0+22i+(22i)}{4}=1+i\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh ya i mean not this, the one wiht just adding

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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
 one year ago
Best ResponseYou've already chosen the best response.2this is the same thing just said a bit differently

dan815
 one year ago
Best ResponseYou've already chosen the best response.2with the 4 vectors 2 at a time share a b c d

Empty
 one year ago
Best ResponseYou've already chosen the best response.1True but your method fails if the square looks like this: dw:1434884123357:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.2well its something thoughj still as long as the point is somethere in the square

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Actually 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we can also just shift the square by some point, so that the origin is located anywhere in the square

Empty
 one year ago
Best ResponseYou've already chosen the best response.1But you can see that there will be uniquely only two numbers for each of the four vectors after translating it to the center.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah but dan do you not understand what I said before?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Like 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
 one year ago
Best ResponseYou've already chosen the best response.2oh ya i can see that one works no matter what

dan815
 one year ago
Best ResponseYou've already chosen the best response.2im just seeing if there is a less computation way in some cases

dan815
 one year ago
Best ResponseYou've already chosen the best response.2not that it matters lol

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I think this is the least computational way unless you wanna do it their gross way lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.2well like (a+b, c+d) is center is slightly faster than (a,c) + (a,d) + (b,d)+ (b,c)/ 4

dan815
 one year ago
Best ResponseYou've already chosen the best response.2only 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
 one year ago
Best ResponseYou've already chosen the best response.1I don't understand how your calculation is realistic though

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I'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
 one year ago
Best ResponseYou've already chosen the best response.1In that example what is a,b,c, and d if there are coefficients 2,1,4,3,5,0 ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ohh i see the problem okayy

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ah okay that makes sense

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ok 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
 one year ago
Best ResponseYou've already chosen the best response.2they should allow rotation in the draw tool, that give so many more options

Empty
 one year ago
Best ResponseYou've already chosen the best response.1If 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
 one year ago
Best ResponseYou've already chosen the best response.2what do you mean reguarly spaced and subtract the center vector