Forcing complex output to take the form $a + b,i$
Forcing complex output to take the form $a + b,i$
Solving the complex equation $z^2=1+2,i$ using
Solve[z^2 == 1 + 2 I]
returns $leftleftzto -sqrt1+2,iright,leftztosqrt1+2,irightright$, but how do I force Mathematica to always output on the form $a+b,i$, $a,binmathbbR$? Or, if there is no output form from Solve to do this, to convert/transform the answer to the $a+b,i$ form?
Solve
I tried
z = a + b I;
Solve[z^2 == 1 + 2 I, a, b ∈ Reals, a, b]
which returns
$$leftleftato-sqrtfrac12left(1+sqrt5right),btosqrtfrac12left(1+sqrt5right)-fracleft(1+sqrt5right)^3/22sqrt2right,leftato sqrtfrac12left(1+sqrt5right),btofracleft(1+sqrt5right)^3/22sqrt2-sqrtfrac12left(1+sqrt5right)rightright$$
but don't think it's a very elegant (and short) way to solve the equation.
One solution is, in its best presentation
$$z_1=sqrtfrac1+sqrt52+isqrtfrac21+sqrt5$$
Can this be output from Solve (or transformation of the output from Solve)?
Solve
Solve[z^2 == 1 + 2 I] // ComplexExpand // FunctionExpand
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Related link: mathematica.stackexchange.com/questions/173930/…
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– mathe
Sep 10 '18 at 2:21
2 Answers
2
Perhaps this?
ComplexExpand[Solve[z^2 == 1 + 2 I], TargetFunctions -> Re, Im]
(*
z -> -5^(1/4) Cos[ArcTan[2]/2] - I 5^(1/4) Sin[ArcTan[2]/2],
z -> 5^(1/4) Cos[ArcTan[2]/2] + I 5^(1/4) Sin[ArcTan[2]/2]
*)
Update:
I saw Bill Watts' comment after I posted my first answer, which suggests FunctionExpand will help with the trig. functions. Simplifying the separate parts as follows gets the result closer to the desired form:
FunctionExpand
FunctionExpand@ComplexExpand[Solve[z^2 == 1 + 2 I]] /.
x_?NumericQ :> ToRadicals@FullSimplify[Re[x]] + I ToRadicals@FullSimplify[Im[x]]
(*
z -> -I Sqrt[1/2 (-1 + Sqrt[5])] - Sqrt[1/2 (1 + Sqrt[5])],
z -> I Sqrt[1/2 (-1 + Sqrt[5])] + Sqrt[1/2 (1 + Sqrt[5])]
*)
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Truly amazing. Will it work on any "simpler" complex equation or is it targeted to a special form of z^2=a+bi equations? How on earth do one learn this syntax?
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– mf67
Sep 10 '18 at 0:16
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@mf67 It should be general, although it's hard to say whether what Mathematica considers simplified will coincide with what is desired. Here are some tutorials on transformation rules: reference.wolfram.com/language/tutorial/…
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– Michael E2
Sep 10 '18 at 0:38
Well, you can define your own simplifying/expanding functions. For example, the following does the trick:
simplify[exp_] := FullSimplify[FunctionExpand[exp], ComplexityFunction -> ((LeafCount[#] + 100 Count[#, (_Root | _Sin | _Cos), 0, Infinity]) &)]
rootExpand[exp_] := exp /. Sqrt[Complex[a_, b_]] :> (simplify[(a^2 + b^2)^(1/4) Cos[1/2 Arg[a + I b]]] + I simplify[(a^2 + b^2)^(1/4) Sin[1/2 Arg[a + I b]]])
Use them as follows:
Solve[z^2 == 1 + 2 I]
% // rootExpand
(* z -> -Sqrt[1 + 2 I], z -> Sqrt[1 + 2 I] *)
(* z -> -I Sqrt[1/2 (-1 + Sqrt[5])] - Sqrt[1/2 (1 + Sqrt[5])], z -> I Sqrt[1/2 (-1 + Sqrt[5])] + Sqrt[1/2 (1 + Sqrt[5])] *)
Solve[z^2 == 7 + 4 I]
% // rootExpand
(* z -> -Sqrt[7 + 4 I], z -> Sqrt[7 + 4 I] *)
(* z -> -I Sqrt[1/2 (-7 + Sqrt[65])] - Sqrt[1/2 (7 + Sqrt[65])], z -> I Sqrt[1/2 (-7 + Sqrt[65])] + Sqrt[1/2 (7 + Sqrt[65])] *)
FWIW: I think your solution is nice too (I don't find it inelegant nor unreasonably long). You can use the function simplify I defined above to make your output slightly nicer looking:
simplify
Solve[(a + I b)^2 == 1 + 2 I, a, b [Element] Reals, a, b] // simplify
(* a -> -Sqrt[1/2 (1 + Sqrt[5])], b -> -Sqrt[1/2 (-1 + Sqrt[5])], a -> Sqrt[1/2 (1 + Sqrt[5])], b -> Sqrt[1/2 (-1 + Sqrt[5])] *)
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Solve[z^2 == 1 + 2 I] // ComplexExpand // FunctionExpandgets you pretty close.$endgroup$
– Bill Watts
Sep 9 '18 at 23:57