If a and b are positive integers and i is the imaginary unit, what is the only real number c for which (a+bi)^4-24i=c?
Expanding (a + bi)4 = a4 + 4a3bi + 6a2b2i2 + 4ab3i3 + b4i4 = a4 + 4a3bi - 6a2b2 - 4ab3i + b4
Since all the i-terms must cancel for the answer to be a real number: 4a3bi - 4ab3i - 24i must have value 0.
---> 4abi(a2 - b2) - 24i = 0
---> ab(a2 - b2) - 6 = 0
---> ab(a2 - b2) = 6
Therefore, using factors of 6: either ab = 1 and a2 - b2 = 6 which is impossible because a = 1 and b = 1 and then a2 - b2 isn't 6
or ab = 2 and a2 - b2 = 3 which means a = 2 and b = 1 -- this works!
or ab = 3 and a2 - b2 = 2 which means a = 3 and b = 2 which is impossible
or ab = 6 and a2 - b2 = 1 which means a = 6 and b = 1 which is impossible
Now that you know that a = 2 and b = 1, place these back into the original formula (a + bi)4 - 24i to find the value of c.
Expanding (a + bi)4 = a4 + 4a3bi + 6a2b2i2 + 4ab3i3 + b4i4 = a4 + 4a3bi - 6a2b2 - 4ab3i + b4
Since all the i-terms must cancel for the answer to be a real number: 4a3bi - 4ab3i - 24i must have value 0.
---> 4abi(a2 - b2) - 24i = 0
---> ab(a2 - b2) - 6 = 0
---> ab(a2 - b2) = 6
Therefore, using factors of 6: either ab = 1 and a2 - b2 = 6 which is impossible because a = 1 and b = 1 and then a2 - b2 isn't 6
or ab = 2 and a2 - b2 = 3 which means a = 2 and b = 1 -- this works!
or ab = 3 and a2 - b2 = 2 which means a = 3 and b = 2 which is impossible
or ab = 6 and a2 - b2 = 1 which means a = 6 and b = 1 which is impossible
Now that you know that a = 2 and b = 1, place these back into the original formula (a + bi)4 - 24i to find the value of c.