+0

# Alg2/geo

0
456
2
+256

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?

May 11, 2019

#1
+21953
+1

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.

May 11, 2019

#1
+21953
+1

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.

geno3141 May 11, 2019
#2
+111455
0

Nice, Geno   !!!

CPhill  May 11, 2019