# Return to Answer

 3 deleted 9 characters in body edited Jul 21 '18 at 3:17 Mr.Wizard♦ 235k3030 gold badges488488 silver badges10901090 bronze badges You have given contradictive assumptions. In Mma the condition that a variable, say, c, is positive (c>0) automatically means that it belongs to Reals. Thus, when you fix c \[Element]∈ Complexes && (a | b | c) > 0 you mislead Mma. According to your initial expressions the parameters a and b are Reals and positive, while c is Complex, am I right? If yes, try this:  expr = Simplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2], Assumptions -> {a, b} > 0]; MapAt[PowerExpand, expr, {2, 1}] (* 1/2 Sqrt[-((I Sqrt[b^2 - 16 Im[c]^2 - 16 Re[c]^2])/a)] *)  Have fun! Edit: To address your question: {2,1} is a TreeCoordinate of the part of the whole expression that is under the outer square root. The Tree you can visualize by the function TreeForm[expr]  yielding the following structure Here the arrow indicates the element {2,1} that wee need. This can be made visible, if you hover the cursor over this element. It is this element that the PowerExpand function is convenient to be applied to. You have given contradictive assumptions. In Mma the condition that a variable, say, c, is positive (c>0) automatically means that it belongs to Reals. Thus, when you fix c \[Element] Complexes && (a | b | c) > 0 you mislead Mma. According to your initial expressions the parameters a and b are Reals and positive, while c is Complex, am I right? If yes, try this:  expr = Simplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2], Assumptions -> {a, b} > 0]; MapAt[PowerExpand, expr, {2, 1}] (* 1/2 Sqrt[-((I Sqrt[b^2 - 16 Im[c]^2 - 16 Re[c]^2])/a)] *)  Have fun! Edit: To address your question: {2,1} is a TreeCoordinate of the part of the whole expression that is under the outer square root. The Tree you can visualize by the function TreeForm[expr]  yielding the following structure Here the arrow indicates the element {2,1} that wee need. This can be made visible, if you hover the cursor over this element. It is this element that the PowerExpand function is convenient to be applied to. You have given contradictive assumptions. In Mma the condition that a variable, say, c, is positive (c>0) automatically means that it belongs to Reals. Thus, when you fix c ∈ Complexes && (a | b | c) > 0 you mislead Mma. According to your initial expressions the parameters a and b are Reals and positive, while c is Complex, am I right? If yes, try this:  expr = Simplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2], Assumptions -> {a, b} > 0]; MapAt[PowerExpand, expr, {2, 1}] (* 1/2 Sqrt[-((I Sqrt[b^2 - 16 Im[c]^2 - 16 Re[c]^2])/a)] *)  Have fun! Edit: To address your question: {2,1} is a TreeCoordinate of the part of the whole expression that is under the outer square root. The Tree you can visualize by the function TreeForm[expr]  yielding the following structure Here the arrow indicates the element {2,1} that wee need. This can be made visible, if you hover the cursor over this element. It is this element that the PowerExpand function is convenient to be applied to. 2 added 546 characters in body edited Feb 9 '18 at 10:51 Alexei Boulbitch 22.9k2727 silver badges7878 bronze badges You have given contradictive assumptions. In Mma the condition that a variable, say, c, is positive (c>0) automatically means that it belongs to Reals. Thus, when you fix c \[Element] Complexes && (a | b | c) > 0 you mislead Mma. According to your initial expressions the parameters a and b are Reals and positive, while c is Complex, am I right? If yes, try this:  expr = Simplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2], Assumptions -> {a, b} > 0]; MapAt[PowerExpand, expr, {2, 1}] (* 1/2 Sqrt[-((I Sqrt[b^2 - 16 Im[c]^2 - 16 Re[c]^2])/a)] *)  Have fun! Edit: To address your question: {2,1} is a TreeCoordinate of the part of the whole expression that is under the outer square root. The Tree you can visualize by the function TreeForm[expr]  yielding the following structure Here the arrow indicates the element {2,1} that wee need. This can be made visible, if you hover the cursor over this element. It is this element that the PowerExpand function is convenient to be applied to. You have given contradictive assumptions. In Mma the condition that a variable, say, c, is positive (c>0) automatically means that it belongs to Reals. Thus, when you fix c \[Element] Complexes && (a | b | c) > 0 you mislead Mma. According to your initial expressions the parameters a and b are Reals and positive, while c is Complex, am I right? If yes, try this:  expr = Simplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2], Assumptions -> {a, b} > 0]; MapAt[PowerExpand, expr, {2, 1}] (* 1/2 Sqrt[-((I Sqrt[b^2 - 16 Im[c]^2 - 16 Re[c]^2])/a)] *)  Have fun! You have given contradictive assumptions. In Mma the condition that a variable, say, c, is positive (c>0) automatically means that it belongs to Reals. Thus, when you fix c \[Element] Complexes && (a | b | c) > 0 you mislead Mma. According to your initial expressions the parameters a and b are Reals and positive, while c is Complex, am I right? If yes, try this:  expr = Simplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2], Assumptions -> {a, b} > 0]; MapAt[PowerExpand, expr, {2, 1}] (* 1/2 Sqrt[-((I Sqrt[b^2 - 16 Im[c]^2 - 16 Re[c]^2])/a)] *)  Have fun! Edit: To address your question: {2,1} is a TreeCoordinate of the part of the whole expression that is under the outer square root. The Tree you can visualize by the function TreeForm[expr]  yielding the following structure Here the arrow indicates the element {2,1} that wee need. This can be made visible, if you hover the cursor over this element. It is this element that the PowerExpand function is convenient to be applied to. 1 answered Feb 9 '18 at 10:11 Alexei Boulbitch 22.9k2727 silver badges7878 bronze badges You have given contradictive assumptions. In Mma the condition that a variable, say, c, is positive (c>0) automatically means that it belongs to Reals. Thus, when you fix c \[Element] Complexes && (a | b | c) > 0 you mislead Mma. According to your initial expressions the parameters a and b are Reals and positive, while c is Complex, am I right? If yes, try this:  expr = Simplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2], Assumptions -> {a, b} > 0]; MapAt[PowerExpand, expr, {2, 1}] (* 1/2 Sqrt[-((I Sqrt[b^2 - 16 Im[c]^2 - 16 Re[c]^2])/a)] *)  Have fun!