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Square Root of Complex Numbers

by littleHilbert
Tags: complex, numbers, root, square
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littleHilbert
#1
Dec25-06, 01:34 PM
P: 57
Hi! I've got a question.
There is a nice formula for finding square roots of arbitrary complex numbers z=a+bi:

[itex]\frac{1}{\sqrt{2}}(\epsilon\sqrt{|z|+a}+i\sqrt{|z|-a})[/itex] where
epsilon:=sing(b) if b≠0 or epsilon:=1 if b=0.

I've just looked it up and it's nice to use it to find complex roots of quadratic equations with complex coefficients.

Where does it come from? I mean, squaring shows that it's true but how can one derive it from other facts? Is there a similar formula for n-th roots (not in polar form but analogous to that above)? Any info, links?
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Amr Morsi
#2
Dec26-06, 09:45 PM
P: 25
LittileHilbrt;

Let y=c+id, where y^2=z.

Then; c^2-d^2+2i c*d=a+ib.

As a result; a=c^2-d^2 (eq1) and b=2c*d (eq2).

Therefore, from eq1 and eq2; c^2 = a + (b/c)^2/4 (eq3);

Solving eq3 in c^2, we get c^2={a+sqr(a^2+b^2)}^.5 /2

Therefore, c=(+/-) sqr({a+mod(z)}/2) (eq4 ),
{the other solution is rejected as c must be real}.

Then; z^1/2 = (+/-) y = (+/-) [c + id] = (+/-) [c+id] = (+/-) [sqr({a+mod(z)})+i sign(b)* sqr({[mod(z)]^2 ľa^2} / {a+mod(z)})]/sqr(2)
= (+/-) [ sqr( a+mod(z) ) + i sign(b) * sqr({mod(z)ľa} ] / sqr(2)
= (+/-) [sign(b) * sqr( a+mod(z) ) + i sqr({mod(z)ľa} ] / [sqr(2)* sign(b)]
====> (+/-) [sign(b) * sqr( a+mod(z) ) + i sqr({mod(z)ľa} ] / [sqr(2)]

Solving these equaitons when b=0 will need some modifications. By the way, when b is zero, z is real.

Amr Morsi.
Werg22
#3
Dec26-06, 10:59 PM
P: 1,520
It can probably be derived with few identities and Demoivre's theorem.

Alkatran
#4
Dec26-06, 11:19 PM
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Square Root of Complex Numbers

You could always convert the number to e^(x + iy) form, divide x and y by 2, convert back to normal form. I think that should work.
Morbius
#5
Dec27-06, 02:23 PM
Sci Advisor
P: 1,152
Quote Quote by littleHilbert View Post
Where does it come from? I mean, squaring shows that it's true but how can one derive it from other facts? Is there a similar formula for n-th roots (not in polar form but analogous to that above)? Any info, links?
littleHilbert,

When you multiply by a complex number, you are rotating the other number
by the angle the complex number makes with the x-axis.

For example, multiplication by "i" makes a rotation of 90 degrees. Therefore
a multiplication by the square of "i" makes a rotation of 180 degrees or -1.
Hence "i" is the square root of -1.

If you want the "n-th root" of a number; express it in polar form, and
divide the angle by "n". You will get the polar expression for the n-th
root. Use trigonometric identities to transform back into the non-polar
form in terms of the components of the original complex number.
Normalize appropriately if the modulus of the number is not unity.

That's how the formula you cite above is derived.

Dr. Gregory Greenman
Physicist
Karlisbad
#6
Dec27-06, 04:41 PM
P: 134
the question is..take the identity:

[tex] (\sqrt (-2)+1)(\sqrt (-2)-1)=-2 [/tex] and expand it by a continous fraction..what would we get..


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