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

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

Mathematics
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how about just Re^itheta = 7+24i
I don't know
hm thats not nice

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|dw:1434880616850:dw|
the angle seems to be exact in degrees
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.
oh thats a nice trick
Next one looks like \[\sqrt{-8-6i}\] I did the same thing basically but it looks like I'm getting two answers.
I'm getting that it's equal to both \[1-3i\] and \[-1+3i\]
there could be 2 answers
going forward and backwards
|dw:1434881253505:dw|
both of them look about right
i wonder if there is another way to do this, like how about normalizing and finding the half angle first
Makes sense since one is the negative of the other, we could just factor out like this: \[\sqrt{-8-6i}=\pm i\sqrt{8+6i}\]
and then sqrting the radius
then convert that to cartesian
or would that get messy with e and pi in that mix
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
I think this ties in nicely with what you were saying earlier, what does the square root of this complex number mean...?
for some reason i always see this anon stalker in my questions all the time http://prntscr.com/7ji57g
Is there a visual way in which we can consider the square root of -8-6i as being 1-3i or -1+3i.
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
I guess that's what you did, you drew it then rotated in either direction
yeah
and then took the square root of the length
right ya squaring i mean not doubling, i confuse those 2
Yeah I gotcha, ok here's a good one looks kinda simple though http://prntscr.com/7ji5ww
we could see they are all orthogonal or
wait we can do that thing from before, normalize all and all the vectors can be written in that e^i2pik/4 form
or just take dot products and that bs
We could subtract one of the vectors from the other 3 to make that point the origin with 3 points coming off it
then we know that one of the vectors must be 90 degrees of the other, so we kinda maybe figure that out and rotate
is there a way to rewrite all these points in a rotated dimension, like a simple way
like if i wanna just rewqrite all these vectors but with the x and y axis rorated to one of the vectors
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?
oh my bad i think i am confusing this question, for some reason i was thinking they are giving us vectors..
they just gave us points ah
Points, vectors, whatever
yea this still works i guess
dot
that will turn them into vectors so we can do this agian
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
|dw:1434882161951:dw|
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|
|dw:1434882255217:dw| Since those are the midpoints on a square, this is a square.
Their solution is ugly as crap: http://www.exampleproblems.com/wiki/index.php/CV19
|dw:1434882335886:dw|
No danny don't break space and do that infinitely otherwise we'll prove that points are really squares
oh thats gross they are writing it out as lines lol
haha
Yeah they're doing it the way I would program a computer to do it maybe, but I have a brain tyvm
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.
why aren't you using complex exponentials first one: \(z = 25 e^{i \theta}, tan \theta = 24/7\) \(\sqrt{z} = \pm 5 e^{i \theta/2}\)
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
this way u can break out of the loop as soon as one if it fails
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.
So 4 equations with 4 unknowns, r, \(\theta\) are just real numbers and z has 2 numbers since it's complex. Coolio.
yeah, that wud be nice, but we wont always get lucky
What do you mean why not?
they can give us annoying points that arent centered
They don't have to be centered for this to work
find we have to take the double diagonal and reposition that intersection as the origin
Woah! I just saw something miraculous
If it's a square then we can add up all the 4 complex numbers to get the center of the square.
wait really
Then we subtract this from the 4 numbers to center it.
no matter where the squre is placed?
It's like vectors in equilibrium.
But the equilibrium is the center displaced
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.
|dw:1434883019314:dw|
Then we can literally square any of the numbers we have to get the other 3, after we normalize them.
|dw:1434883081383:dw|
yaa thats what i mean though, u wont get nice vertices though
watch, I'll do it.
ohh wait! ur saying additiong of these 4 vectors will give u center, i can see it kind of :O
woah thats cool, it makes sense it has to go toward the center
Yeah this is the best way for a computer to do it too for any polygon.
the farther away from center u are the more u are pulled to the otehr vertices
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.
this gives us a much simpler way of solving the square question now
this works for all regular polygons
Yeah
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.
Wait this doesn't quite work hahaha
It should be the average of the points not just the sum of the points.
ya i was just looking at that lol
|dw:1434883586373:dw|
\[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\] Subtract that from each of those same 4 points: -2i, 2, 2i, -2.
What could be sexier?
|dw:1434883777297:dw|
like thats an easy one to see that fails for sure
No, that doesn't fail
why not
\[\frac{0+2-2i+(2-2i)}{4}=1+i\]
oh ya i mean not this, the one wiht just adding
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.
|dw:1434883967181:dw|
this is the same thing just said a bit differently
with the 4 vectors 2 at a time share a b c d
True but your method fails if the square looks like this: |dw:1434884123357:dw|
oh truee
well its something thoughj still as long as the point is somethere in the square
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.
|dw:1434884274366:dw|
lol
No offense....
lolol
we can also just shift the square by some point, so that the origin is located anywhere in the square
But you can see that there will be uniquely only two numbers for each of the four vectors after translating it to the center.
Yeah but dan do you not understand what I said before?
which one?
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.
oh ya i can see that one works no matter what
im just seeing if there is a less computation way in some cases
not that it matters lol
I think this is the least computational way unless you wanna do it their gross way lol
well like (-a+b, -c+d) is center is slightly faster than (a,c) + (a,d) + (b,d)+ (b,c)/ 4
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
I don't understand how your calculation is realistic though
why not
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.\[\frac{(2+i)+(4+3i)+(2+5i)+3i}{4} = 2+3i\]
In that example what is a,b,c, and d if there are coefficients 2,1,4,3,5,0 ?
ohh i see the problem okayy
ah okay that makes sense
the rotations
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
they should allow rotation in the draw tool, that give so many more options
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 ;)
what do you mean reguarly spaced and subtract the center vector
|dw:1434885527859:dw|
that vector?
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.
ah okay that makes sense
once u subtract the center vector every point is share by 2 vectors
every componnent of the point
wait no
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
are u sure this formula works
a square at the center unrotated vs rotated would have tme same number of unique vector components with that formula, but it shouldnt
does it have to more a multiple of 4 more than times 1
have to be*
|dw:1434886153355:dw|
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
or actually 4 at a time
No, rethink this, you're on the right track though.
|dw:1434886388250:dw|