If (x+iy)^3 = −74 + ki, find the absolute value of k, given that x=1 and i^2 =−1.
(x + iy)^3 = -74 + ki
x^3 + 3x^2*iy + 3x * (iy)^2 + ( iy)^3 = -74 + ki
(1)^3 + 3*yi + 3y^2i^2 + (yi)^3 = -74 + ki
1 + 3yi - 3y^2 - y^3i = -74 + ki
- 3y^2 + (3yi - y^3i) = -75 + ( ki )
Equate
-3y^2 = -75
y^2 = -75/-3
y^2 = 25
y = 5 or y = -5
If y = 5....then If y = -5......then
3(5)i - (5^3)i = ki 3 (-5)i - (-5)^3 i = ki
15i - 125i = ki -15i + 125i = ki
-110 i = ki 110i = ki
-110 = k 110 = k