What is ³√(1+i)?

.

z = ³√(1 + i)

z³ = 1 + i

　 = √2[ (1/√2) + i•(1/√2) ]

　 = √2[cos45 + i•sin45] , √2[cos405 + i•sin405] , √2[cos765 + i•sin765]

z = ⁶√2[cos15 + i•sin15] , ⁶√2[cos135 + i•sin135] , ⁶√2[cos255 + i•sin255]

　= ⁶√2[cos15 + i•sin15] , ⁶√2[cos135 + i•sin135] , ⁶√2[- cos75 - i•sin75]

z₁ = ⁶√2[ (√6 + √2)/4 + i•(√6 - √2)/4]

z₂ = ⁶√2[ (- 1/√2) + i•(1/√2)]

z₃ = ⁶√2[- (√6 - √2)/4 - i•(√6 + √2)/4]

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³√(1+i) is right.

http://www.wolframalpha.com/input/?i=%C2%B3%E2%88%...

Well,

as is, the notation ³√( . ) is reserved for the cubic root and the domain of this function is : R+ = [0, +oo)

so : ³√(1+i) does not make sense, as there are several complex numbers z so that z^3 = 1+i !! ;-)

now what are these numbers ?

we have :

1 + i = sqrt2(cos(π/4) + i sin(π/4) )

= 2^(1/2) e^(i *π/4) using Euler's notation

therefore :

the roots are given by :

zk = 2^(1/6 )* e^(i *π/12) + 2i kπ/12) for k = 0, 1, 2

hope it' ll help !!

i+1 = √2 e^(i pi/4)

Therefore

³√(1+i) = (i+1)^(1/3) = (√2)^(1/3) e^(i pi/12)

You can convert to rectangular (i.e. x+ iy) notation if you like.

Bear in mind that every complex number has three cube roots, separated by 120° on a circle centered at the origin.

z = a + ib

The modulus is: m = √(a² + b²)

The argument is such as: tan(β) = b/a

z = √(1 + i)

The modulus is: m = √(1² + 1²) = √2

The argument is such as: tan(β) = 1/1 → β = π/4

You want to get: z = ³√(1 + i)

The modulus is: = ³√m = ³√(√2)

The argument is: = (β/3) = π/12