There are a few arctangent identities that are useful for these sorts of problems. In this case, the identity atan(11/2) = 3 * atan(1/2) may be used. Just convert 2 + i*11 to polar form, then use the identity to take one third of the exponent, and the real cube root of the modulus, and then you are almost done. Just take the sin and cos of atan(1/2), to get the root 2 + i. To find all three cube roots, multiply that by exp(i*k*2*pi/3) for k = 0, 1, 2 Here are the most commonly useful arctan identities for these kinds of problems: atan(11/2) = 3 * atan(1/2) atan(13/9) = 3 * atan(1/3) atan(46/9) = pi - 3 * atan(2/3) Also, recall that atan(1/x) = pi/2 - atan(x) There are quite a few obscure tan and atan identities floating around.
Yes, this is easy if you know the identity arctan(¹¹⁄₂) = 3·arctan(¹⁄₂) but what if you don't? Let's say that (1) 2 + 11i = r(cos φ + i·sin φ) = r·exp(i·φ) with r ≥ 0, −π < φ ≤ π then r is the modulus of 2 + 11i and φ is its principal argument. The principal cube root of 2 + 11i is then (2) ∛(2 + 11i) = ∛r·(cos ⅓φ + i·sin ⅓φ) = ∛r·exp(i·⅓φ) Finding ∛r is easy enough, since r² = 2² + 11² = 4 + 121 = 125 so r = √125 = 5√5 and so ∛r = √5. From (1) we also easily find cos φ = 2/r = 2/(5√5) = ²⁄₂₅√5 and sin φ = 11/r = 11/(5√5) = ¹¹⁄₂₅√5. We now need to find cos ⅓φ and sin ⅓φ and we could do this by using the triple angle identities of the cosine and the sine and solve for cos ⅓φ and sin ⅓φ from the known values of cos φ and sin φ, but this requires solving two cubic equations. Alternatively, we can start from tan φ = sin φ/cos φ = 11/2 and since 2 + 11i is in the right half of the complex plane and therefore −½π < φ < ½π the principal argument of 2 + 11i is φ = arctan(11/2). We then need to find a real value x such that ⅓φ = arctan(x) so (3) 3·arctan(x) = arctan(11/2) WolframAlpha seems uncapable to provide an exact solution of this equation in any usable form, but it is possible to solve (3) exactly. From φ = arctan(11/2) and ⅓φ = arctan(x) we have tan φ = 11/2 and tan ⅓φ = x and using the triple angle identity tan 3α = (3·tan α − tan³α)/(1 − 3·tan²α) with α = ⅓φ we have (4) (3x − x³)/(1 − 3x²) = 11/2 which gives (5) 2x³ − 33x² − 6x + 11 = 0 Note that this is exactly the same cubic equation as the equation in t obtained in the video at 3:50. Equation (5) has _three_ real solutions x = ½, x = 8 − 5√3, x = 8 + 5√3 and these obviously all satisfy (6) tan(3·arctan(x)) = (11/2) But since arctan(x) is strictly increasing on ℝ equation (3) can only have _one_ real solution. From φ = arctan(11/2) we know that φ > 0 and since we also have −½π < φ < ½π it follows that 0 < φ < ½π and therefore 0 < ⅓φ < ⅙π so 0 < tan ⅓φ < tan ⅙π, that is, 0 < x < ⅓√3, and x = ½ is the only root of (5) which satisfies this condition, so x = ½ is the only solution of (3), meaning that we have (7) 3·arctan(1/2) = arctan(11/2) From tan ⅓φ = x = ½ we have 1/cos²(⅓φ) = 1 + tan²(⅓φ) = 1 + ¼ = ⁵⁄₄ so cos²(⅓φ) = ⅘ which gives cos ⅓φ = 2/√5 = ⅖√5 since −½π < φ < ½π implies cos ⅓φ > 0. In turn this gives sin ⅓φ = tan ⅓φ·cos ⅓φ = ½·⅖√5 = ⅕√5. Finally, Substituting ∛r = √5, cos ⅓φ = ⅖√5, sin ⅓φ = ⅕√5 in (2) we find that the principal cube root of 2 + 11i is (8) ∛(2 + 11i) = √5·(⅖√5 + i·⅕√5) = 2 + i The two other (non principal) cube roots of 2 + 11i can of course be obtained by multiplying the principal value 2 + i with the complex cube roots of unity ω₁ = − ½ + ½i√3 and ω₂ = −½ − ½i√3 which gives (− ½ + ½i√3)(2 + i) = (−1 − ½√3) + (−½ + √3)i (− ½ − ½i√3)(2 + i) = (−1 + ½√3) + (−½ − √3)i Alternatively, noting that the tangent function has a period π, we can write the three solutions of (6) and therefore also the three solutions of (5) as (9a) x = tan(⅓·arctan(¹¹⁄₂)) (9b) x = tan(⅓·(arctan(¹¹⁄₂) − π)) (9c) x = tan(⅓·(arctan(¹¹⁄₂) + π)) and from this and the respective solutions x = ½, x = 8 − 5√3, x = 8 + 5√3 we get (10a) 3·arctan(½) = arctan(¹¹⁄₂) (10b) 3·arctan(8 − 5√3) = arctan(¹¹⁄₂) − π (10c) 3·arctan(8 + 5√3) = arctan(¹¹⁄₂) + π and we can write the three cube roots of 2 + 11i in exponential form as √5·exp(i·arctan(½)) = 2 + i √5·exp(i·(arctan(8 − 5√3) + π)) = (−1 − ½√3) + (−½ + √3)i √5·exp(i·(arctan(8 + 5√3) − π)) = (−1 + ½√3) + (−½ − √3)i This looks horrific, but you can verify with a calculator that it is correct.
Your equation using t is basically the triple angle formula for cotan. We can see the modulus of the complex expression is sqrt(125) so its cube root is sqrt(5). The argument is arctan(11/2)/3, which you get from the triple angle formula.
Method 2 on me. Express 2+11i in polar form. Modulus would be 5 sqrt5 and the argument theta is given by tan inverse of 11/2. The cubic root shall have modulus of cubic root of 5sqrt5 which ends up being sqrt5. The argument would be theta/3. Using the triple angle formula tan theta/3 would be 1/2. First quadrant solution would give the principal root. Other two roots are given by simply adding 2PI/3 and 4PI/3 to the argument of the principal root respectively.
A slightly different approach consists in realizing that if we have two real numbers a and b such that (1) (a + bi)³ = 2 + 11i then we must also have (2) (a − bi)³ = 2 − 11i Multiplying (1) and (2) using the identity (a + bi)(a − bi) = a² + b² and (2 + 11i)(2 − 11i) = 4 − (−121) = 125 = 5³ we have (3) a² + b² = 5 since a and be are real. Expanding the left hand side of (1) and equating the real and imaginary parts we have (4a) a³ − 3ab² = 2 (4b) 3a²b − b³ = 11 From (3) we get b² = 5 − a² and a² = 5 − b² and substituting the first in (4a) and the second in (4b) we get (5a) 4a³ − 15a − 2 = 0 (5b) 4b³ − 15b + 11 = 0 Using the rational root theorem we can find that a = 2 is a root of (5a) and that b = 1 is a root of (5b) which allows us to factor (5a) as (a − 2)(4a² + 8a + 1) = 0 and (5b) as (b − 1)(4b² + 4b − 11) = 0 and this allows us to easily find all roots of (5a) and (5b) which are (6a) a = 2 ⋁ a = −1 + ½√3 ⋁ a = −1 − ½√3 (6b) b = 1 ⋁ b = −½ + √3 ⋁ b = −½ − √3 So, there are three possible real values of a and three possible real values of b. But we cannot freely combine these to obtain nine complex numbers whose cube is 2 + 11i because a and b must also satisfy (3). There are three combinations for a + bi such that a² + b² = 5 and these are 2 + i (−1 − ½√3) + (−½ + √3)i (−1 + ½√3) + (−½ − √3)i The cube of each of these complex numbers is 2 + 11i but there can be only one _principal_ cube root of 2 + 11i, which is 2 + i. The principal value of the cube root ∛z of a complex number z = r·exp(φi) with r ≥ 0, −π < φ ≤ π is defined as ∛z = ∛r·exp(⅓φi) where ∛r is the nonnegative real cube root of the nonnegative real number r.
The book is _Imagining Numbers (particularly the square root of minus fifteen)_ by Barry Mazur and the title refers to Cardano's finding that two numbers with sum 10 and product 40 could be expressed as 5 + √−15 and 5 − √−15. In his 1545 book _Ars Magna_ (the great art) which is the first publication in which algebraic solutions of cubic and quartic equations are discussed (although Cardano did not discover the solution methods) he describes the result (5 + √−15)(5 − √−15) = 25 − (−15) = 25 + 15 = 40 But he couldn't really get his head around this, because in describing this he uses the words _dimissis incruciationibus_ which is usually translated as 'putting aside the mental tortures involved', but there may be a play of words here because this can also be understood as 'the cross-multiples having canceled out'.
Thanks a lot, you save me a lot of time,it cames to my mind how to solve this kind of problems,I have similar one,they just switched real part to imaginary part.
Hi I would like to add some theorem about the problem of nrooth. Theorem: The nrooth of complex number Z is exactly the product of one of the rooth multiply by nrooth of unity. Sorry if the translation is not exact from french
There are a few arctangent identities that are useful for these sorts of problems. In this case, the
identity atan(11/2) = 3 * atan(1/2) may be used. Just convert 2 + i*11 to polar form, then use the identity to take one third of the exponent, and the real cube root of the modulus, and then you are almost done. Just take the sin and cos of atan(1/2), to get the root 2 + i. To find all three cube roots, multiply that by exp(i*k*2*pi/3) for k = 0, 1, 2
Here are the most commonly useful arctan identities for these kinds of problems:
atan(11/2) = 3 * atan(1/2)
atan(13/9) = 3 * atan(1/3)
atan(46/9) = pi - 3 * atan(2/3)
Also, recall that atan(1/x) = pi/2 - atan(x)
There are quite a few obscure tan and atan identities floating around.
Yes, this is easy if you know the identity arctan(¹¹⁄₂) = 3·arctan(¹⁄₂) but what if you don't? Let's say that
(1) 2 + 11i = r(cos φ + i·sin φ) = r·exp(i·φ) with r ≥ 0, −π < φ ≤ π
then r is the modulus of 2 + 11i and φ is its principal argument. The principal cube root of 2 + 11i is then
(2) ∛(2 + 11i) = ∛r·(cos ⅓φ + i·sin ⅓φ) = ∛r·exp(i·⅓φ)
Finding ∛r is easy enough, since r² = 2² + 11² = 4 + 121 = 125 so r = √125 = 5√5 and so ∛r = √5. From (1) we also easily find cos φ = 2/r = 2/(5√5) = ²⁄₂₅√5 and sin φ = 11/r = 11/(5√5) = ¹¹⁄₂₅√5.
We now need to find cos ⅓φ and sin ⅓φ and we could do this by using the triple angle identities of the cosine and the sine and solve for cos ⅓φ and sin ⅓φ from the known values of cos φ and sin φ, but this requires solving two cubic equations.
Alternatively, we can start from tan φ = sin φ/cos φ = 11/2 and since 2 + 11i is in the right half of the complex plane and therefore −½π < φ < ½π the principal argument of 2 + 11i is φ = arctan(11/2). We then need to find a real value x such that ⅓φ = arctan(x) so
(3) 3·arctan(x) = arctan(11/2)
WolframAlpha seems uncapable to provide an exact solution of this equation in any usable form, but it is possible to solve (3) exactly. From φ = arctan(11/2) and ⅓φ = arctan(x) we have tan φ = 11/2 and tan ⅓φ = x and using the triple angle identity tan 3α = (3·tan α − tan³α)/(1 − 3·tan²α) with α = ⅓φ we have
(4) (3x − x³)/(1 − 3x²) = 11/2
which gives
(5) 2x³ − 33x² − 6x + 11 = 0
Note that this is exactly the same cubic equation as the equation in t obtained in the video at 3:50. Equation (5) has _three_ real solutions x = ½, x = 8 − 5√3, x = 8 + 5√3 and these obviously all satisfy
(6) tan(3·arctan(x)) = (11/2)
But since arctan(x) is strictly increasing on ℝ equation (3) can only have _one_ real solution. From φ = arctan(11/2) we know that φ > 0 and since we also have −½π < φ < ½π it follows that 0 < φ < ½π and therefore 0 < ⅓φ < ⅙π so 0 < tan ⅓φ < tan ⅙π, that is, 0 < x < ⅓√3, and x = ½ is the only root of (5) which satisfies this condition, so x = ½ is the only solution of (3), meaning that we have
(7) 3·arctan(1/2) = arctan(11/2)
From tan ⅓φ = x = ½ we have 1/cos²(⅓φ) = 1 + tan²(⅓φ) = 1 + ¼ = ⁵⁄₄ so cos²(⅓φ) = ⅘ which gives cos ⅓φ = 2/√5 = ⅖√5 since −½π < φ < ½π implies cos ⅓φ > 0. In turn this gives sin ⅓φ = tan ⅓φ·cos ⅓φ = ½·⅖√5 = ⅕√5. Finally, Substituting ∛r = √5, cos ⅓φ = ⅖√5, sin ⅓φ = ⅕√5 in (2) we find that the principal cube root of 2 + 11i is
(8) ∛(2 + 11i) = √5·(⅖√5 + i·⅕√5) = 2 + i
The two other (non principal) cube roots of 2 + 11i can of course be obtained by multiplying the principal value 2 + i with the complex cube roots of unity ω₁ = − ½ + ½i√3 and ω₂ = −½ − ½i√3 which gives
(− ½ + ½i√3)(2 + i) = (−1 − ½√3) + (−½ + √3)i
(− ½ − ½i√3)(2 + i) = (−1 + ½√3) + (−½ − √3)i
Alternatively, noting that the tangent function has a period π, we can write the three solutions of (6) and therefore also the three solutions of (5) as
(9a) x = tan(⅓·arctan(¹¹⁄₂))
(9b) x = tan(⅓·(arctan(¹¹⁄₂) − π))
(9c) x = tan(⅓·(arctan(¹¹⁄₂) + π))
and from this and the respective solutions x = ½, x = 8 − 5√3, x = 8 + 5√3 we get
(10a) 3·arctan(½) = arctan(¹¹⁄₂)
(10b) 3·arctan(8 − 5√3) = arctan(¹¹⁄₂) − π
(10c) 3·arctan(8 + 5√3) = arctan(¹¹⁄₂) + π
and we can write the three cube roots of 2 + 11i in exponential form as
√5·exp(i·arctan(½)) = 2 + i
√5·exp(i·(arctan(8 − 5√3) + π)) = (−1 − ½√3) + (−½ + √3)i
√5·exp(i·(arctan(8 + 5√3) − π)) = (−1 + ½√3) + (−½ − √3)i
This looks horrific, but you can verify with a calculator that it is correct.
Your equation using t is basically the triple angle formula for cotan.
We can see the modulus of the complex expression is sqrt(125) so its cube root is sqrt(5). The argument is arctan(11/2)/3, which you get from the triple angle formula.
Method 2 on me.
Express 2+11i in polar form. Modulus would be 5 sqrt5 and the argument theta is given by tan inverse of 11/2.
The cubic root shall have modulus of cubic root of 5sqrt5 which ends up being sqrt5.
The argument would be theta/3. Using the triple angle formula tan theta/3 would be 1/2. First quadrant solution would give the principal root.
Other two roots are given by simply adding 2PI/3 and 4PI/3 to the argument of the principal root respectively.
Nice
That was a clever way to find all three roots. I liked it.
Can you please solve z^3=√3 - i√3
A slightly different approach consists in realizing that if we have two real numbers a and b such that
(1) (a + bi)³ = 2 + 11i
then we must also have
(2) (a − bi)³ = 2 − 11i
Multiplying (1) and (2) using the identity (a + bi)(a − bi) = a² + b² and (2 + 11i)(2 − 11i) = 4 − (−121) = 125 = 5³ we have
(3) a² + b² = 5
since a and be are real. Expanding the left hand side of (1) and equating the real and imaginary parts we have
(4a) a³ − 3ab² = 2
(4b) 3a²b − b³ = 11
From (3) we get b² = 5 − a² and a² = 5 − b² and substituting the first in (4a) and the second in (4b) we get
(5a) 4a³ − 15a − 2 = 0
(5b) 4b³ − 15b + 11 = 0
Using the rational root theorem we can find that a = 2 is a root of (5a) and that b = 1 is a root of (5b) which allows us to factor (5a) as (a − 2)(4a² + 8a + 1) = 0 and (5b) as (b − 1)(4b² + 4b − 11) = 0 and this allows us to easily find all roots of (5a) and (5b) which are
(6a) a = 2 ⋁ a = −1 + ½√3 ⋁ a = −1 − ½√3
(6b) b = 1 ⋁ b = −½ + √3 ⋁ b = −½ − √3
So, there are three possible real values of a and three possible real values of b. But we cannot freely combine these to obtain nine complex numbers whose cube is 2 + 11i because a and b must also satisfy (3). There are three combinations for a + bi such that a² + b² = 5 and these are
2 + i
(−1 − ½√3) + (−½ + √3)i
(−1 + ½√3) + (−½ − √3)i
The cube of each of these complex numbers is 2 + 11i but there can be only one _principal_ cube root of 2 + 11i, which is 2 + i. The principal value of the cube root ∛z of a complex number z = r·exp(φi) with r ≥ 0, −π < φ ≤ π is defined as ∛z = ∛r·exp(⅓φi) where ∛r is the nonnegative real cube root of the nonnegative real number r.
❤
I took an approach that WA probably took, but nice job!
reminds me of a book entitled something like "Imaginary numbers and the cube root of minus 15".
The book is _Imagining Numbers (particularly the square root of minus fifteen)_ by Barry Mazur and the title refers to Cardano's finding that two numbers with sum 10 and product 40 could be expressed as 5 + √−15 and 5 − √−15. In his 1545 book _Ars Magna_ (the great art) which is the first publication in which algebraic solutions of cubic and quartic equations are discussed (although Cardano did not discover the solution methods) he describes the result
(5 + √−15)(5 − √−15) = 25 − (−15) = 25 + 15 = 40
But he couldn't really get his head around this, because in describing this he uses the words _dimissis incruciationibus_ which is usually translated as 'putting aside the mental tortures involved', but there may be a play of words here because this can also be understood as 'the cross-multiples having canceled out'.
Thanks a lot, you save me a lot of time,it cames to my mind how to solve this kind of problems,I have similar one,they just switched real part to imaginary part.
Can you write formula of (a+bi)^n
Sure
Hi I would like to add some theorem about the problem of nrooth.
Theorem:
The nrooth of complex number Z is exactly the product of one of the rooth multiply by nrooth of unity.
Sorry if the translation is not exact from french