To solve this expression, we need to first square the given complex number.
Let z = 2(cos(11π/12) + i sin(11π/12))
Now, z^2 = [2(cos(11π/12) + i sin(11π/12))]^2= 4[cos(11π/12) + i sin(11π/12)]^2= 4[cos(22π/12) + i sin(22π/12)]= 4[cos(11π/6) + i sin(11π/6)]= 4[√3/2 + i * 1/2]= 2√3 + 2i
Therefore, the result is 2√3 + 2i.
To solve this expression, we need to first square the given complex number.
Let z = 2(cos(11π/12) + i sin(11π/12))
Now, z^2 = [2(cos(11π/12) + i sin(11π/12))]^2
= 4[cos(11π/12) + i sin(11π/12)]^2
= 4[cos(22π/12) + i sin(22π/12)]
= 4[cos(11π/6) + i sin(11π/6)]
= 4[√3/2 + i * 1/2]
= 2√3 + 2i
Therefore, the result is 2√3 + 2i.