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October 20, 2009 05:05 PM
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We have i = cos(pi/2) + i.sin(pi/2)
sqrt(i) = (cos(pi/2) + i.sin(pi/2))^(1/2)
By DeMoivre's theorem:
= cos(pi/4) + i.sin(pi/4)
1 + i
= ----------
sqrt(2)
To learn more about DeMoivre and his theorem, look at:
http://mathforum.org/library/drmath/view/53975.html
you can find another explanation here
http://www.math.toronto.edu/mathnet/questionCorner/rootofi.html
Source(s):
http://www.math.toronto.edu/mathnet/questionCorner/rootofi.htm
http://mathforum.org/library/drmath/view/53975.htm
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To calculate the square root you need to take the square root of the magnitude (in this case sqrt(1) = 1), and half the angle (in this case 90 degrees divided by 2 = 45 degrees). The result is thus a vector of magnitude 1 pointing half way between the positive Real direction and the positive Imaginary direction.
This is 1/sqrt(2) + i/sqrt(2).
However, the angle could also be half of -270 degrees, or -135 degrees.
This would result in -1/sqrt(2) - i/sqrt(2).
An easy way to see why is to realize that this second solution is simply the first one, multiplied by -1. Thus, taking the square of the second solution gives you (-1)^2 multiplied by the first answer squared. Since (-1)^2 = 1, you get the same result by squaring either solution.
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Then sqrt(i) = a + ai.
(a + ai)(a + ai) = a^2 + 2a^2i -a"2 = 2 * 1/2 * i = i
This video explains complex number multiplication in geometrical terms, which show explains why the answer comes out to be what it does....
http://www.youtube.com/watch?v=S7NDSPtoVP0
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take square root on both sides twice, one gets "a" square root of i as
e**(pi/4 * i)
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What is the square root of i?
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Answers (6)
October 20, 2009 05:27 PM
The square root of i We have i = cos(pi/2) + i.sin(pi/2)
sqrt(i) = (cos(pi/2) + i.sin(pi/2))^(1/2)
By DeMoivre's theorem:
= cos(pi/4) + i.sin(pi/4)
1 + i
= ----------
sqrt(2)
To learn more about DeMoivre and his theorem, look at:
http://mathforum.org/library/drmath/view/53975.html
you can find another explanation here
http://www.math.toronto.edu/mathnet/questionCorner/rootofi.html
Source(s):
http://www.math.toronto.edu/mathnet/questionCorner/rootofi.htm
http://mathforum.org/library/drmath/view/53975.htm
Permalink | Report
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October 20, 2009 05:29 PM
To calculate this you need to understand how roots work in the complex plain. This is the 2-D plain with the Real number axis along the horizontal and the Imaginary number axis along the vertical. In this plane, i is a vector of length 1 along the positive Imaginary direction (i.e. angle of +90 degrees). To calculate the square root you need to take the square root of the magnitude (in this case sqrt(1) = 1), and half the angle (in this case 90 degrees divided by 2 = 45 degrees). The result is thus a vector of magnitude 1 pointing half way between the positive Real direction and the positive Imaginary direction.
This is 1/sqrt(2) + i/sqrt(2).
However, the angle could also be half of -270 degrees, or -135 degrees.
This would result in -1/sqrt(2) - i/sqrt(2).
An easy way to see why is to realize that this second solution is simply the first one, multiplied by -1. Thus, taking the square of the second solution gives you (-1)^2 multiplied by the first answer squared. Since (-1)^2 = 1, you get the same result by squaring either solution.
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October 20, 2009 05:45 PM
If a = sqrt(1/2) = 0.707106.... Then sqrt(i) = a + ai.
(a + ai)(a + ai) = a^2 + 2a^2i -a"2 = 2 * 1/2 * i = i
This video explains complex number multiplication in geometrical terms, which show explains why the answer comes out to be what it does....
http://www.youtube.com/watch?v=S7NDSPtoVP0
Permalink | Report
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October 21, 2009 08:19 PM
This is ia well known theorem e**(pi*i) = -1 take square root on both sides twice, one gets "a" square root of i as
e**(pi/4 * i)
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Voted as best: thehero1989
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"i" represents an - or perhaps, THE - IMAGINARY NUMBER!!! I followed your link, ikilian, and... Whoa, now my brain started to hurt some more.
But, then, I looked around the Dr. Math (mathforum.org) site some more, and found an FAQ on "imaginary numbers." That made it a bit more clear. (http://mathforum.org/dr.math/faq/faq.imag.num.html )
But THEN, I clicked on the explanation for ELEMENTARY SCHOOL STUDENTS... and I understand!!!
That explanation doesn't contain any trig functions, btw. So, I still can't figure out whether any of the answers here match the one that Dr. Math gives. But, check out the explanation, and the answer to the square root of "i" is at the very bottom. (It looks like he or she is saying that the answer to chazzyfen's question is sqrt of 2 over 2 PLUS sqrt of 2 over 2 times "i". iklilian, doesn't that mean that sqrt of 3i would be 2.1210 + 2.1210i?)
Wow, my brain doesn't hurt as much, but that is definitely some weird stuff!!! Thanks to all for helping increase my body of knowledge1
(Here's the link to the elem school level explanation: http://mathforum.org/library/drmath/view/58730.html)