(√2+i√3)×i√6−√2×(2i√2−3)+i×(6√6−6√3)=x−y+z
I am going to divide it into 3 parts.
x = (sqrt2 + i sqrt3) * i sqrt6
y = sqrt2 *(2i sqrt2 - 3)
z = i * (6/sqrt6 - 6/sqrt3)
x = (√2+i√3)×i√6=i√12+i2√18=2i√3−3√2
y = √2×(2i√2−3)=4i−3√2
z = i×(6√6−6√3)=i×(6−6√2√6)=i×(√6−√12)=i√6−2i√3
x - y + z
=(2i√3−2√2)−(4i−3√2)+(i√6−2i√3)=2i√3−2√2−4i+3√2+i√6−2i√3=√2+i(√6−4)
Therefore the original equation is sqrt2 + i(sqrt6 - 4)
Simplify the following:
(sqrt(2)+i sqrt(3))×i sqrt(6)-sqrt(2) (2 i sqrt(2)-3)+i (6/sqrt(6)-(6)/sqrt(3))
Rationalize the denominator. 6/sqrt(6) = 6/sqrt(6)×(sqrt(6))/(sqrt(6)) = (6 sqrt(6))/(6):
(sqrt(2)+i sqrt(3))×i sqrt(6)-sqrt(2) (2 i sqrt(2)-3)+i ((6 sqrt(6))/(6)-(6)/sqrt(3))
(6 sqrt(6))/(6) = 6/6×sqrt(6) = sqrt(6):
(sqrt(2)+i sqrt(3))×i sqrt(6)-sqrt(2) (2 i sqrt(2)-3)+i (sqrt(6)-(6)/sqrt(3))
Rationalize the denominator. (-6)/sqrt(3) = (-6)/sqrt(3)×(sqrt(3))/(sqrt(3)) = (-6 sqrt(3))/(3):
(sqrt(2)+i sqrt(3))×i sqrt(6)-sqrt(2) (2 i sqrt(2)-3)+i (sqrt(6)+(-6 sqrt(3))/(3))
(-6)/3 = (3 (-2))/3 = -2:
(sqrt(2)+i sqrt(3))×i sqrt(6)-sqrt(2) (2 i sqrt(2)-3)+i (sqrt(6)+-2 sqrt(3))
sqrt(6) (sqrt(2)+i sqrt(3)) = 3 i sqrt(2)+2 sqrt(3):
3 i sqrt(2)+2 sqrt(3) i-sqrt(2) (2 i sqrt(2)-3)+i (sqrt(6)-2 sqrt(3))
i (3 i sqrt(2)+2 sqrt(3)) = 2 i sqrt(3)-3 sqrt(2):
2 i sqrt(3)-3 sqrt(2)-sqrt(2) (2 i sqrt(2)-3)+i (sqrt(6)-2 sqrt(3))
-(2 i sqrt(2)-3) = 3+-2 i sqrt(2):
-3 sqrt(2)+2 i sqrt(3)+sqrt(2) 3-2 i sqrt(2)+i (sqrt(6)-2 sqrt(3))
sqrt(2) (3+-2 i sqrt(2)) = -4 i+3 sqrt(2):
-3 sqrt(2)+2 i sqrt(3)+-4 i+3 sqrt(2)+i (sqrt(6)-2 sqrt(3))
Add like terms. -3 sqrt(2)+2 i sqrt(3)-4 i+3 sqrt(2)+i (sqrt(6)-2 sqrt(3)) = -4 i+2 i sqrt(3)+i (sqrt(6)-2 sqrt(3)):
-4 i+2 i sqrt(3)+i (sqrt(6)-2 sqrt(3))
Factor i from -4 i+2 i sqrt(3)+i (sqrt(6)-2 sqrt(3)):
i (-4+2 sqrt(3)-2 sqrt(3)+sqrt(6))
Add like terms. -4+2 sqrt(3)-2 sqrt(3)+sqrt(6) = sqrt(6)-4:
Answer: |i (sqrt(6) - 4)