Complex Number - Real part of Sqrt(1+z)?
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From: torteloni on 30 Mar 2010 15:25 Hey Group, how can I calculate the real part (and later on the imaginery as well) o the following expression (z being an arbitrary complex number): Sqrt(1+z) Derive would get me the solution of a+bi=Sqrt(1+c+di) quickly (with i being the imaginery unit and hence c+di=z): a=1/2 Sqrt[2] Sqrt[Sqrt[c^2 + 2c + d^2 + 1] + c + 1] and b=1/2 Sqrt[2] Sqrt[Sqrt[c^2 + 2c + d^2 + 1] - c - 1] Sign[d] (box brackets because I copied it from Mathematica). Which makes sense but anyway I am highly interested in how to solve it by hand. Thank you very much for your help. Daniel
From: Rod on 30 Mar 2010 16:18 "torteloni" <torteloni(a)googlemail.com> wrote in message news:ee92d8f0-752f-4310-9c9c-d5f70bb0c279(a)z11g2000yqz.googlegroups.com... > Hey Group, > > how can I calculate the real part (and later on the imaginery as well) > o the following expression (z being an arbitrary complex number): > > Sqrt(1+z) > > Derive would get me the solution of a+bi=Sqrt(1+c+di) quickly (with i > being the imaginery unit and hence c+di=z): > > a=1/2 Sqrt[2] Sqrt[Sqrt[c^2 + 2c + d^2 + 1] + c + 1] > > and > > b=1/2 Sqrt[2] Sqrt[Sqrt[c^2 + 2c + d^2 + 1] - c - 1] Sign[d] > > (box brackets because I copied it from Mathematica). Which makes sense > but anyway I am highly interested in how to solve it by hand. > > Thank you very much for your help. > > Daniel > any complex number can be written x+iy= r exp(i theta). r real = sqrt(x^2+y^2) tan theta = y/z so the sqrt becomes sqrt(r) exp(i theta/2) Of course there are two sqrts, so where is the other one. well exp(i 2*pi) = 1 so can always write exp(i theta) = 1* exp(i theta) = exp( i (theta + 2pi)) so hte sqrt becomes exp ( i*(theta + 2pi)/2) = exp(i theta/2 + i*pi) exp(i*pi) = -1 so find the other root. This last bit gets to be a bit more important when looking at larger roots like cube roots then you get a term like exp(i*2pi/3) which is not quite so trivial
From: Dave L. Renfro on 30 Mar 2010 17:17 torteloni wrote (in part): > how can I calculate the real part (and later on the imaginery > as well) o[f] the following expression (z being an arbitrary > complex number): > > Sqrt(1+z) Rod has posted one way. Here's another way, one that only relies on high school algebra. Start with sqrt(1 + z) = sqrt(1 + x + iy) = U + iV, where U and V are the real and imaginary parts you want. Square both sides. 1 + x + iy = (U^2 - V^2) + (2UV)i Equating real and imaginary parts gives you (1) U^2 - V^2 = 1 + x (2) 2UV = y Now just solve this quadratic system for U and V in terms of x and y. First, note that if y = 0, then (2) implies that U = 0 or V = 0, and hence with (1) we get U^2 = 1 + x or -V^2 = 1 + x. Because U^2 and V^2 are non-negative real-valued expressions (recall U and V are real-valued expressions), we get: 1 + x >= 0 implies U = sqrt(1+x) and V = 0 1 + x <= 0 implies U = 0 and V = i*sqrt(-1 - x). Of course, this is what we should get if you look back to the original equation sqrt(1 + x + iy) = U + iV. Now assume y is nonzero. Since 2UV = y, we get V = y/(2U), and so (1) becomes U^2 - y^2/(4U^2) = 1 + x 4U^4 - y^2 = 4(1+x)U^2 4U^4 - 4(1+x)U^2 - y^2 = 0 This is quadratic in U^2, so we can use the quadratic formula: U^2 = { 4(1+x) +/- sqrt[16(1+x)^2 + 16y^2] } / 8 U^2 = { 4(1+x) +/- 4*sqrt[(1+x)^2 + y^2] } / 8 U^2 = { (1+x) +/- sqrt[(1+x)^2 + y^2] } / 2 Since y is nonzero, the negative sign choice leads to U^2 being negative (recall y is a real number, so y^2 would be positive, hence sqrt[(1+x)^2 + y^2] would be a positive value larger in magnitude than 1+x, hence U^2 would be negative; note that my argument works whether 1+x is positive, negative, or zero), which is a contradiction, since by definition of "real part", U is a real-valued expression. Therefore, we have U^2 = {(1+x) + sqrt[(1+x)^2 + y^2]} / 2 U = +/- sqrt{ {(1+x) + sqrt[(1+x)^2 + y^2]} / 2 } As for V, just use V = y/(2U). Note that there are two solution-pairs for (U,V). Dave L. Renfro
From: Ken Pledger on 30 Mar 2010 17:33 In article <ee92d8f0-752f-4310-9c9c-d5f70bb0c279(a)z11g2000yqz.googlegroups.com>, torteloni <torteloni(a)googlemail.com> wrote: > .... > how can I calculate the real part (and later on the imaginery as well) > o the following expression (z being an arbitrary complex number): > > Sqrt(1+z) > .... If this is from a text-book, then I don't think much of the author's elementary but laborious problem. Anyway, here's something you can do with it. Write sqrt(1 + z) = x + iy, square both sides, then equate the real and imaginary parts. That should give you equations for x^2 - y^2 and 2xy. Eliminating y gives an equation which is quadratic in x^2, so you can then slog out the rest. Afterwards you'll really deserve your cup of coffee or whatever. Ken Pledger.
From: Rob Johnson on 30 Mar 2010 17:44
In article <ee92d8f0-752f-4310-9c9c-d5f70bb0c279(a)z11g2000yqz.googlegroups.com>, torteloni <torteloni(a)googlemail.com> wrote: >how can I calculate the real part (and later on the imaginery as well) >o the following expression (z being an arbitrary complex number): > >Sqrt(1+z) > >Derive would get me the solution of a+bi=Sqrt(1+c+di) quickly (with i >being the imaginery unit and hence c+di=z): > >a=1/2 Sqrt[2] Sqrt[Sqrt[c^2 + 2c + d^2 + 1] + c + 1] > >and > >b=1/2 Sqrt[2] Sqrt[Sqrt[c^2 + 2c + d^2 + 1] - c - 1] Sign[d] > >(box brackets because I copied it from Mathematica). Which makes sense >but anyway I am highly interested in how to solve it by hand. Perhaps this formula will help sqrt(x^2 + y^2) + x sqrt(x + iy) = sqrt( ------------------- ) 2 sqrt(x^2 + y^2) - x + i sign(y) sqrt( ------------------- ) 2 and of course the negative of the expression above is the other square root. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |