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bonez
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12 May 2009, 3:17 pm

I know I ask a lot of math questions here.. Trigonometry is soooo complicated! Thanks everyone for all of your help.

ok so i have to find the complex square root of -3 + 3*sqr(3i)
first i have to covert it to polar form and then find the square root using the formula.
so i know that i have to put it in a + bi format and then use a and b to find r ( r = sqr(a^2 + b ^2).
so i think a = -3 and i dont know how to find be because of the square root. am i doing all the steps correctly so far? and how do i finish the rest?
Thanks alot!



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12 May 2009, 3:25 pm

Has anyone taught you about complex logarithms. Might help you to know that log(z) = log |z| +i arg z and also that a^b = e^(b log a) under complex logarithms.

So to find our complex square root, start by finding the modulus and argument (constrained to the principal branch -pi < z < pi ). Then use the above rules carefully to get the answer, remembering that a square root is merely raising the complex number by the power of a half.



bonez
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12 May 2009, 3:31 pm

but dont i need to convert to polar form first? how do u convert to polar form if part of it is in square root form?



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12 May 2009, 3:37 pm

Well, first of all, if your number is written in rectangular coordinates, I think you mean that
z=-3+i(3√3) (well, the i could be under the radical, but that complicates this problem considerably).
Hence, a=-3, b=3√3.
Once you have R, you should know z=Re^[i(θ+2nπ)].
You therefore need to find a θ satisfying the cosθ and sinθ; this is basic trig. From there, it's just a matter of basic rules of exponentiation.

And complex logarithms are entirely unnecessary.


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Kangoogle
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12 May 2009, 3:55 pm

twoshots wrote:
Well, first of all, if your number is written in rectangular coordinates, I think you mean that
z=-3+i(3√3) (well, the i could be under the radical, but that complicates this problem considerably).
Hence, a=-3, b=3√3.
Once you have R, you should know z=Re^[i(θ+2nπ)].
You therefore need to find a θ satisfying the cosθ and sinθ; this is basic trig. From there, it's just a matter of basic rules of exponentiation.

And complex logarithms are entirely unnecessary.

Assuming he has got the question stated wrong, then yes. Otherwise complex logs are neccessary and imo are something a 19-yo doing maths should know.



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12 May 2009, 4:12 pm

Well, no. The method you gave is basically the same as mine; writing a^b=e^(b loga) is tantamount to just writing it in polar form, but that's just a straightforward consequence of Euler's Formula, which he's significantly more likely to be familiar with than complex logs and which is more fundamental.


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12 May 2009, 10:08 pm

Code:
z = (-3 + 3*(3i)^(1/2))^(1/2)

= 3^(1/2) (-1 + (3i)^(1/2))^(1/2)

g := (3i)^(1/2)

g = 3^(1/2) i^(1/2)

= 3^(1/2) (± (2^(1/2)/2 (i + 1))

g = 6^(1/2)/2 i + 6^(1/2)/2

z_1 = 3^(1/2) (-1 + g)^(1/2)

z_1 = 3^(1/2) (-1 + 6^(1/2)/2 - 6^(1/2)/2 i)^(1/2)

z_1r = 3^(1/2) (1 + ((4 - 3 6^(1/2) + 6)^(1/2) - 6^(1/2)/2)/2)^(1/2)
z_1i = ± 3^(1/2) (- 1 + ((4 - 3 6^(1/2) + 6)^(1/2) + 6^(1/2)/2)/2)^(1/2)


For z_2 analog.



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13 May 2009, 10:16 am

bonez wrote:
ok so i have to find the complex square root of -3 + 3*sqr(3i)


Google -3 + 3*sqr(3i) returns (-3) + (3 * sqr(3 * i)) = 0.674234614 + 3.67423461 i

:)


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Kangoogle
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13 May 2009, 10:44 am

twoshots wrote:
Well, no. The method you gave is basically the same as mine; writing a^b=e^(b loga) is tantamount to just writing it in polar form, but that's just a straightforward consequence of Euler's Formula, which he's significantly more likely to be familiar with than complex logs and which is more fundamental.

Someone doing maths at age 19 should know the basics of complex analysis, I certainly learned all about that sort of stuff at that age. In fact I have an exam on it all in a couple of weeks.