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Simplify \(\sqrt{7+24i}\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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dan815
  • dan815
how about just Re^itheta = 7+24i
Empty
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I don't know
dan815
  • dan815
hm thats not nice

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dan815
  • dan815
|dw:1434880616850:dw|
dan815
  • dan815
the angle seems to be exact in degrees
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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
  • dan815
oh thats a nice trick
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Next one looks like \[\sqrt{-8-6i}\] I did the same thing basically but it looks like I'm getting two answers.
Empty
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I'm getting that it's equal to both \[1-3i\] and \[-1+3i\]
dan815
  • dan815
there could be 2 answers
dan815
  • dan815
going forward and backwards
dan815
  • dan815
|dw:1434881253505:dw|
dan815
  • dan815
both of them look about right
dan815
  • dan815
i wonder if there is another way to do this, like how about normalizing and finding the half angle first
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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
  • dan815
and then sqrting the radius
dan815
  • dan815
then convert that to cartesian
dan815
  • dan815
or would that get messy with e and pi in that mix
Empty
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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
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I think this ties in nicely with what you were saying earlier, what does the square root of this complex number mean...?
dan815
  • dan815
for some reason i always see this anon stalker in my questions all the time http://prntscr.com/7ji57g
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Is there a visual way in which we can consider the square root of -8-6i as being 1-3i or -1+3i.
dan815
  • 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
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I guess that's what you did, you drew it then rotated in either direction
dan815
  • dan815
yeah
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and then took the square root of the length
dan815
  • dan815
right ya squaring i mean not doubling, i confuse those 2
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Yeah I gotcha, ok here's a good one looks kinda simple though http://prntscr.com/7ji5ww
dan815
  • dan815
we could see they are all orthogonal or
dan815
  • 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
  • dan815
or just take dot products and that bs
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We could subtract one of the vectors from the other 3 to make that point the origin with 3 points coming off it
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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
  • dan815
is there a way to rewrite all these points in a rotated dimension, like a simple way
dan815
  • dan815
like if i wanna just rewqrite all these vectors but with the x and y axis rorated to one of the vectors
Empty
  • 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
  • dan815
oh my bad i think i am confusing this question, for some reason i was thinking they are giving us vectors..
dan815
  • dan815
they just gave us points ah
Empty
  • Empty
Points, vectors, whatever
dan815
  • dan815
yea this still works i guess
dan815
  • dan815
dot
dan815
  • dan815
that will turn them into vectors so we can do this agian
dan815
  • 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
  • dan815
|dw:1434882161951:dw|
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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|
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|dw:1434882255217:dw| Since those are the midpoints on a square, this is a square.
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Their solution is ugly as crap: http://www.exampleproblems.com/wiki/index.php/CV19
dan815
  • dan815
|dw:1434882335886:dw|
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No danny don't break space and do that infinitely otherwise we'll prove that points are really squares
dan815
  • dan815
oh thats gross they are writing it out as lines lol
dan815
  • dan815
haha
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Yeah they're doing it the way I would program a computer to do it maybe, but I have a brain tyvm
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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
  • 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
  • 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
  • dan815
this way u can break out of the loop as soon as one if it fails
Empty
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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.
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So 4 equations with 4 unknowns, r, \(\theta\) are just real numbers and z has 2 numbers since it's complex. Coolio.
dan815
  • dan815
yeah, that wud be nice, but we wont always get lucky
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What do you mean why not?
dan815
  • dan815
they can give us annoying points that arent centered
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They don't have to be centered for this to work
dan815
  • dan815
find we have to take the double diagonal and reposition that intersection as the origin
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Woah! I just saw something miraculous
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If it's a square then we can add up all the 4 complex numbers to get the center of the square.
dan815
  • dan815
wait really
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Then we subtract this from the 4 numbers to center it.
dan815
  • dan815
no matter where the squre is placed?
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It's like vectors in equilibrium.
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But the equilibrium is the center displaced
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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
  • dan815
|dw:1434883019314:dw|
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Then we can literally square any of the numbers we have to get the other 3, after we normalize them.
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|dw:1434883081383:dw|
dan815
  • dan815
yaa thats what i mean though, u wont get nice vertices though
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watch, I'll do it.
dan815
  • dan815
ohh wait! ur saying additiong of these 4 vectors will give u center, i can see it kind of :O
dan815
  • dan815
woah thats cool, it makes sense it has to go toward the center
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Yeah this is the best way for a computer to do it too for any polygon.
dan815
  • dan815
the farther away from center u are the more u are pulled to the otehr vertices
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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
  • dan815
this gives us a much simpler way of solving the square question now
dan815
  • dan815
this works for all regular polygons
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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
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Wait this doesn't quite work hahaha
Empty
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It should be the average of the points not just the sum of the points.
dan815
  • dan815
ya i was just looking at that lol
dan815
  • dan815
|dw:1434883586373:dw|
Empty
  • 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.
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What could be sexier?
dan815
  • dan815
|dw:1434883777297:dw|
dan815
  • dan815
like thats an easy one to see that fails for sure
Empty
  • Empty
No, that doesn't fail
dan815
  • dan815
why not
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\[\frac{0+2-2i+(2-2i)}{4}=1+i\]
dan815
  • 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
  • dan815
|dw:1434883967181:dw|
dan815
  • dan815
this is the same thing just said a bit differently
dan815
  • dan815
with the 4 vectors 2 at a time share a b c d
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True but your method fails if the square looks like this: |dw:1434884123357:dw|
dan815
  • dan815
oh truee
dan815
  • dan815
well its something thoughj still as long as the point is somethere in the square
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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.
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|dw:1434884274366:dw|
dan815
  • dan815
lol
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No offense....
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lolol
dan815
  • dan815
we can also just shift the square by some point, so that the origin is located anywhere in the square
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But you can see that there will be uniquely only two numbers for each of the four vectors after translating it to the center.
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Yeah but dan do you not understand what I said before?
dan815
  • dan815
which one?
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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
  • dan815
oh ya i can see that one works no matter what
dan815
  • dan815
im just seeing if there is a less computation way in some cases
dan815
  • dan815
not that it matters lol
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I think this is the least computational way unless you wanna do it their gross way lol
dan815
  • dan815
well like (-a+b, -c+d) is center is slightly faster than (a,c) + (a,d) + (b,d)+ (b,c)/ 4
dan815
  • 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
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I don't understand how your calculation is realistic though
dan815
  • dan815
why not
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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.\[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\]
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In that example what is a,b,c, and d if there are coefficients 2,1,4,3,5,0 ?
dan815
  • dan815
ohh i see the problem okayy
dan815
  • dan815
ah okay that makes sense
dan815
  • dan815
the rotations
dan815
  • 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
  • dan815
they should allow rotation in the draw tool, that give so many more options
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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
  • dan815
what do you mean reguarly spaced and subtract the center vector
dan815
  • dan815
|dw:1434885527859:dw|
dan815
  • dan815
that vector?
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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
  • dan815
ah okay that makes sense
dan815
  • dan815
once u subtract the center vector every point is share by 2 vectors
dan815
  • dan815
every componnent of the point
dan815
  • dan815
wait no
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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
  • dan815
are u sure this formula works
dan815
  • 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
  • dan815
does it have to more a multiple of 4 more than times 1
dan815
  • dan815
have to be*
dan815
  • dan815
|dw:1434886153355:dw|
dan815
  • 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
  • dan815
or actually 4 at a time
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No, rethink this, you're on the right track though.
dan815
  • dan815
|dw:1434886388250:dw|
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Ahhh ok yeah you're right it is only every 4 I was thinking every 8 I see
dan815
  • dan815
i was cutting green peppers.. and i wiped my eyes
dan815
  • dan815
ok ya i see now
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rofl
dan815
  • dan815
i can see how u ll start start to have rotated vector components
dan815
  • dan815
of pi/2 and -pi/2 and showing up
dan815
  • dan815
so that means a,b --- > -b,a
dan815
  • dan815
and a,b --- > b,-a
dan815
  • dan815
for multiples of 4 the 90 degree ones must exist for every point
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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
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ok nvm that's stupid why did I ask that lol
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http://prntscr.com/7jiujx Let's do this one
dan815
  • dan815
hey if u draw the tangents to 8 polygon u get 2 squares?
dan815
  • dan815
|dw:1434886974869:dw|