
#1
Dec2506, 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: [latex]\frac{1}{\sqrt{2}}(\epsilon\sqrt{z+a}+i\sqrt{za})[/latex] 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 nth roots (not in polar form but analogous to that above)? Any info, links? 



#2
Dec2606, 09:45 PM

P: 25

LittileHilbrt;
Let y=c+id, where y^2=z. Then; c^2d^2+2i c*d=a+ib. As a result; a=c^2d^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 –a^2} / {a+mod(z)})]/sqr(2) = (+/) [ sqr( a+mod(z) ) + i sign(b) * sqr({mod(z)–a} ] / sqr(2) = (+/) [sign(b) * sqr( a+mod(z) ) + i sqr({mod(z)–a} ] / [sqr(2)* sign(b)] ====> (+/) [sign(b) * sqr( a+mod(z) ) + i sqr({mod(z)–a} ] / [sqr(2)] Solving these equaitons when b=0 will need some modifications. By the way, when b is zero, z is real. Amr Morsi. 



#3
Dec2606, 10:59 PM

P: 1,520

It can probably be derived with few identities and Demoivre's theorem.




#4
Dec2606, 11:19 PM

Sci Advisor
HW Helper
P: 944

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.




#5
Dec2706, 02:23 PM

Sci Advisor
P: 1,137

When you multiply by a complex number, you are rotating the other number by the angle the complex number makes with the xaxis. 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 "nth root" of a number; express it in polar form, and divide the angle by "n". You will get the polar expression for the nth root. Use trigonometric identities to transform back into the nonpolar 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 



#6
Dec2706, 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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