Answer on Question #52986 – Math– Complex Analysis
Question:
What is the square root of 2 x + i ( x 2 − 1 ) \sqrt{2x + i(x^2 - 1)} 2 x + i ( x 2 − 1 )
Explanation:
It should be noted that, as it may seem, a complicated way of calculation yields the simple algebraic form of solution. Moreover, if we square the received solution, then we get the initial expression under the square root.
Let's compare the algebraic and the exponential form of solution. By definition, the exponential form of complex number is
Z = X + i Y = ρ e i φ , ρ = ∣ Z ∣ = X 2 + Y 2 , φ = A r g Z = arctan ( Y X ) + 2 π k , k = 0 , ± 1 , ± 2 , … Z = X + iY = \rho e^{i\varphi}, \quad \rho = |Z| = \sqrt{X^2 + Y^2}, \qquad \varphi = \mathrm{Arg}Z = \arctan\left(\frac{Y}{X}\right) + 2\pi k, \; k = 0, \pm 1, \pm 2, \ldots Z = X + iY = ρ e i φ , ρ = ∣ Z ∣ = X 2 + Y 2 , φ = Arg Z = arctan ( X Y ) + 2 πk , k = 0 , ± 1 , ± 2 , …
In our case
Z = 2 x + i ( x 2 − 1 ) , X = 2 x , Y = ( x 2 − 1 ) , Z = 2x + i(x^2 - 1), \qquad X = 2x, Y = (x^2 - 1), Z = 2 x + i ( x 2 − 1 ) , X = 2 x , Y = ( x 2 − 1 ) ,
and we get
Z = ( ρ e i φ ) 1 2 = ρ e i φ 2 , \sqrt{Z} = \left(\rho e^{i\varphi}\right)^{\frac{1}{2}} = \sqrt{\rho} e^{\frac{i\varphi}{2}}, Z = ( ρ e i φ ) 2 1 = ρ e 2 i φ , ρ = ( 2 x ) 2 + ( x 2 − 1 ) 2 = 4 x 2 + x 4 − 2 x 2 + 1 = x 4 + 2 x 2 + 1 = ( x 2 + 1 ) 2 , \rho = \sqrt{(2x)^2 + (x^2 - 1)^2} = \sqrt{4x^2 + x^4 - 2x^2 + 1} = \sqrt{x^4 + 2x^2 + 1} = \sqrt{(x^2 + 1)^2}, ρ = ( 2 x ) 2 + ( x 2 − 1 ) 2 = 4 x 2 + x 4 − 2 x 2 + 1 = x 4 + 2 x 2 + 1 = ( x 2 + 1 ) 2 , φ = arctan ( ( x 2 − 1 ) 2 x ) + 2 π k , \varphi = \arctan\left(\frac{(x^2 - 1)}{2x}\right) + 2\pi k, φ = arctan ( 2 x ( x 2 − 1 ) ) + 2 πk , Z = ( x 2 + 1 ) 2 4 e i 2 ( arctan ( ( x 2 − 1 ) 2 x ) + 2 π k ) = x 2 + 1 e i 2 ( arctan ( ( x 2 − 1 ) 2 x ) + 2 π k ) , k = 0 , ± 1 , ± 2 , … . \sqrt{Z} = \sqrt[4]{(x^2 + 1)^2} e^{\frac{i}{2} \left(\arctan\left(\frac{(x^2 - 1)}{2x}\right) + 2\pi k\right)} = \sqrt{x^2 + 1} e^{\frac{i}{2} \left(\arctan\left(\frac{(x^2 - 1)}{2x}\right) + 2\pi k\right)}, \; k = 0, \pm 1, \pm 2, \ldots. Z = 4 ( x 2 + 1 ) 2 e 2 i ( a r c t a n ( 2 x ( x 2 − 1 ) ) + 2 πk ) = x 2 + 1 e 2 i ( a r c t a n ( 2 x ( x 2 − 1 ) ) + 2 πk ) , k = 0 , ± 1 , ± 2 , … .
We can see the exponential form of solution is bulky and implicit, unlike the algebraic form
Z = ± 1 2 ( ( x + 1 ) + i ( x − 1 ) ) . \sqrt{Z} = \pm \frac{1}{\sqrt{2}} \big((x + 1) + i(x - 1)\big). Z = ± 2 1 ( ( x + 1 ) + i ( x − 1 ) ) . Solution
Let's put
2 x + i ( x 2 − 1 ) = a + i b , \sqrt{2x + i(x^2 - 1)} = a + ib, 2 x + i ( x 2 − 1 ) = a + ib ,
where a , b ∈ R a, b \in \mathbb{R} a , b ∈ R
2 x + i ( x 2 − 1 ) = ( a + i b ) 2 , ⇒ 2x + i(x^2 - 1) = (a + ib)^2, \Rightarrow 2 x + i ( x 2 − 1 ) = ( a + ib ) 2 , ⇒ 2 x + i ( x 2 − 1 ) = ( a 2 − b 2 ) + 2 i a b . 2x + i(x^2 - 1) = (a^2 - b^2) + 2iab. 2 x + i ( x 2 − 1 ) = ( a 2 − b 2 ) + 2 iab .
Comparing the left-hand and right-hand sides of equality (2), we get the following system of equations
{ a 2 − b 2 = 2 x , 2 a b = x 2 − 1. \left\{ \begin{array}{l} a ^ {2} - b ^ {2} = 2 x, \\ 2 a b = x ^ {2} - 1. \end{array} \right. { a 2 − b 2 = 2 x , 2 ab = x 2 − 1.
Let's solve this system. By performing elementary transformations of (3)
{ a 2 − b 2 = 2 x , 2 a b = x 2 − 1 , ⇒ { a 2 − b 2 = 2 x , a = x 2 − 1 2 b , ⇒ ( x 2 − 1 2 b ) 2 − b 2 = 2 x , ⇒ ( x 2 − 1 ) 2 − 4 b 4 = 8 x b 2 , \left\{ \begin{array}{l} a ^ {2} - b ^ {2} = 2 x, \\ 2 a b = x ^ {2} - 1, \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \boxed {a ^ {2} - b ^ {2} = 2 x}, \\ \boxed {a = \frac {x ^ {2} - 1}{2 b}}, \end{array} \right. \Rightarrow \left(\frac {x ^ {2} - 1}{2 b}\right) ^ {2} - b ^ {2} = 2 x, \Rightarrow (x ^ {2} - 1) ^ {2} - 4 b ^ {4} = 8 x b ^ {2}, { a 2 − b 2 = 2 x , 2 ab = x 2 − 1 , ⇒ ⎩ ⎨ ⎧ a 2 − b 2 = 2 x , a = 2 b x 2 − 1 , ⇒ ( 2 b x 2 − 1 ) 2 − b 2 = 2 x , ⇒ ( x 2 − 1 ) 2 − 4 b 4 = 8 x b 2 ,
we receive a biquadratic equation
4 b 4 + 8 x b 2 − ( x 2 − 1 ) 2 = 0. 4 b ^ {4} + 8 x b ^ {2} - (x ^ {2} - 1) ^ {2} = 0. 4 b 4 + 8 x b 2 − ( x 2 − 1 ) 2 = 0.
Its solutions are
b 1 , 2 = ± 1 2 ( x − 1 ) , b 3 , 4 = ± i 2 ( x + 1 ) . b _ {1, 2} = \pm \frac {1}{\sqrt {2}} (x - 1), b _ {3, 4} = \pm \frac {i}{\sqrt {2}} (x + 1). b 1 , 2 = ± 2 1 ( x − 1 ) , b 3 , 4 = ± 2 i ( x + 1 ) .
Further, substituting (5) into a = x 2 − 1 2 b a = \frac{x^2 - 1}{2b} a = 2 b x 2 − 1
a 1 , 2 = x 2 − 1 2 ⋅ ( ± 1 2 ( x − 1 ) ) = ± ( x − 1 ) ( x + 1 ) 2 ( x − 1 ) = ± ( x + 1 ) 2 ; a _ {1, 2} = \frac {x ^ {2} - 1}{2 \cdot \left(\pm \frac {1}{\sqrt {2}} (x - 1)\right)} = \pm \frac {(x - 1) (x + 1)}{\sqrt {2} (x - 1)} = \pm \frac {(x + 1)}{\sqrt {2}}; a 1 , 2 = 2 ⋅ ( ± 2 1 ( x − 1 ) ) x 2 − 1 = ± 2 ( x − 1 ) ( x − 1 ) ( x + 1 ) = ± 2 ( x + 1 ) ; a 3 , 4 = x 2 − 1 2 ⋅ ( ± i 2 ( x + 1 ) ) = ± ( x − 1 ) ( x + 1 ) i 2 ( x + 1 ) = ∓ i ( x − 1 ) 2 ; a _ {3, 4} = \frac {x ^ {2} - 1}{2 \cdot \left(\pm \frac {i}{\sqrt {2}} (x + 1)\right)} = \pm \frac {(x - 1) (x + 1)}{i \sqrt {2} (x + 1)} = \mp \frac {i (x - 1)}{\sqrt {2}}; a 3 , 4 = 2 ⋅ ( ± 2 i ( x + 1 ) ) x 2 − 1 = ± i 2 ( x + 1 ) ( x − 1 ) ( x + 1 ) = ∓ 2 i ( x − 1 ) ;
Therefore, the substituting (5), (6a) and (6b) into (1) yields the following result:
2 x + i ( x 2 − 1 ) = { ± ( x + 1 ) 2 ± i 2 ( x − 1 ) = ± 1 2 ( ( x + 1 ) + i ( x − 1 ) ) ; ∓ i ( x − 1 ) 2 ± i 2 2 ( x + 1 ) = ∓ i ( x − 1 ) 2 ∓ 1 2 ( x + 1 ) = ∓ 1 2 ( ( x + 1 ) + i ( x − 1 ) ) . [ 2 x + i ( x 2 − 1 ) = ± 1 2 ( ( x + 1 ) + i ( x − 1 ) ) ] \begin{array}{l} \sqrt {2 x + i (x ^ {2} - 1)} = \left\{ \begin{array}{c} \pm \frac {(x + 1)}{\sqrt {2}} \pm \frac {i}{\sqrt {2}} (x - 1) = \pm \frac {1}{\sqrt {2}} \big ((x + 1) + i (x - 1) \big); \\ \mp \frac {i (x - 1)}{\sqrt {2}} \pm \frac {i ^ {2}}{\sqrt {2}} (x + 1) = \mp \frac {i (x - 1)}{\sqrt {2}} \mp \frac {1}{\sqrt {2}} (x + 1) = \mp \frac {1}{\sqrt {2}} \big ((x + 1) + i (x - 1) \big). \end{array} \right. \\ \left[ \sqrt {2 x + i \left(x ^ {2} - 1\right)} = \pm \frac {1}{\sqrt {2}} \left((x + 1) + i (x - 1)\right) \right] \\ \end{array} 2 x + i ( x 2 − 1 ) = { ± 2 ( x + 1 ) ± 2 i ( x − 1 ) = ± 2 1 ( ( x + 1 ) + i ( x − 1 ) ) ; ∓ 2 i ( x − 1 ) ± 2 i 2 ( x + 1 ) = ∓ 2 i ( x − 1 ) ∓ 2 1 ( x + 1 ) = ∓ 2 1 ( ( x + 1 ) + i ( x − 1 ) ) . [ 2 x + i ( x 2 − 1 ) = ± 2 1 ( ( x + 1 ) + i ( x − 1 ) ) ]
Answer: 2 x + i ( x 2 − 1 ) = ± 1 2 ( ( x + 1 ) + i ( x − 1 ) ) . \sqrt{2x + i(x^2 - 1)} = \pm \frac{1}{\sqrt{2}}\big((x + 1) + i(x - 1)\big). 2 x + i ( x 2 − 1 ) = ± 2 1 ( ( x + 1 ) + i ( x − 1 ) ) .
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