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Roots of complex numbers (1 Viewer)

wogboy

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Here's something similar to J0n's method, but with a magic shortcut:

(a + ib)^2 = i
a^2 - b^2 + i(2ab) = i
so a^2 - b^2 = 0 AND 2ab = 1

therefore a^2 + b^2 = 2*b^2.
however a^2 + b^2 = |a + ib|^2 = |(a+ib)^2| = |i| = 1
1 = 2*b^2
b = +- 1/sqrt(2)
subbing b into the equation 2ab = 1
so a = +- 1/sqrt(2)

so the square roots of i are:

1/sqrt(2) + i/sqrt(2)
-1/sqrt(2) - i/sqrt(2)
 

KeypadSDM

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Originally posted by abdooooo!!!
hey anyone know if there is a calculator for calculating roots of complex numbers? shouldn't be that hard to do for a calculator... it makes life a lot easier. :)
Find the argument and moduli, and do the rest on your calculator. Halve the angle, square root the moduli, put a +- out the front, and you've got yourself a quick answer. (Note, if your answer doesn't give you rational numbers, just square the components to see what they are).
 

abdooooo!!!

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Originally posted by KeypadSDM
Find the argument and moduli, and do the rest on your calculator. Halve the angle, square root the moduli, put a +- out the front, and you've got yourself a quick answer. (Note, if your answer doesn't give you rational numbers, just square the components to see what they are).
umm... i was looking for a faster way than that... but anyway thats proly the faster it can be done without some sought of special calculator that can not be use in exams.

do any of you people know that google is actually a calculator? try it... type something like "square root of 6+9i" in the search and see the results fast. :D :D :D
 

J0n

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Originally posted by KeypadSDM
How about mathematica. They've got an online integrator, maybe they've got something to do complex calculations as well?
Isn't that what the mathematica software is for??
 

ND

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Originally posted by wogboy
Here's something similar to J0n's method, but with a magic shortcut:

(a + ib)^2 = i
a^2 - b^2 + i(2ab) = i
so a^2 - b^2 = 0 AND 2ab = 1

therefore a^2 + b^2 = 2*b^2.
however a^2 + b^2 = |a + ib|^2 = |(a+ib)^2| = |i| = 1
1 = 2*b^2
b = +- 1/sqrt(2)
subbing b into the equation 2ab = 1
so a = +- 1/sqrt(2)

so the square roots of i are:

1/sqrt(2) + i/sqrt(2)
-1/sqrt(2) - i/sqrt(2)
Hey GuardiaN, does this look familiarto you? (It's that method you were asking me about yesturday)

How about mathematica. They've got an online integrator, maybe they've got something to do complex calculations as well?
I've got Maple 7, it looks like a powerful mathematical program. Too bad i've got absolutely no idea how to use it...
 

ND

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Cool, thanks for that. Do you think that learning maple is useful for uni maths? (i'll be doing actuarial studies, and maybe combining it with maths)
 

wogboy

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Yep it's very useful for uni maths, I had to learn Maple for uni assesments (I'm doing electrical engineering). The actuarial students at UNSW have to learn MATLAB instead though. Not sure about Macquarie uni though, maybe you can ask one of the actuary students on this board. Still it's a good idea to learn Maple, since it might make it easier to then learn other programs like MATLAB etc, once you learn it Maple is quite easy to use.
 

ND

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Ah i see, i know that at macq they learn excel, but is that the equivalent? or would that be used for something different?
 

wogboy

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Excel is quite different, it's a spreadsheet package rather than mathematical software. Excel is kind of more suited to statistical analysis and financial budgeting sort of stuff (e.g. finding means, variance, standard deviations and simple calculations from data sets and displaying graphs etc), while Maple or MATLAB is more for the technical side of maths (e.g. solving differential equations, vectors & matrices, calculus, graphing, and LOTS more advanced stuff)
 

Grey Council

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Berlooody oath, is Wogboy good or what? :)

I was looking for that method (kinda the same thing). Someone else described it in a post, but I didn't quite get it, so I asked Keypad and ND to explain it to me. I get it now. :) Only thing is, it seems to be only useful when you have whole numbers.

eg:
sqrt(3 + 4i) = z
.: x^2 - y^2 = 3 (equating real with real after expanding)
and x^2 + y^2 = 5 (using that 'magical' method)
so obviously x = 2 and y=1 (take negative for other value).

useful, as I said, but not really very useful when the answer ISN'T a whole number. I think.

And you are a genius Wogboy. Seriously. humph

Applying the magical method to sqrt of i:
sqrt(i) = x + iy
x^2 - y^2 = 0
x^2 + y^2 = 1
lol, didn't think of doing x^2=y^2, the way wogboy did it. Thats quite neat, actually. Let me try it here, but my solution will be identical to Wogboy's now.
x^2 = y^2
2*x^2 = 1
x^2 = 1/2
x = 1/(sqrt 2)

Day-YUM! That is neat. very elegant. ffs, why couldn't I have thought of that. :(
 
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J0n

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Originally posted by abdooooo!!!
hey anyone know if there is a calculator for calculating roots of complex numbers? shouldn't be that hard to do for a calculator... it makes life a lot easier. :)
I know a few people at school who have calculators which work with complex numbers - problem is they don't know how to use them. :D
 

Grey Council

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are those calculators allowed? It defeats the whole purpose of complex numbers if they are.

Btw, look at Lazarus' post. ffs, I don't get a word he said.

"Jeeee-ZZZAAA! What a mind you must have"
;) In this case, what a mind he must have.
 

ND

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My calculator [that i used in the exam] has complex numbers on it, and yeh i don't know how to use it either, all i know how to do is convert between rectangular and mod-arg representations (not that i used it for this at all).
 

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