40 35
-
2+i 2-i
How do I solve this?
there is a division sign between the 40 and 2=i and another one between 35 and 2-i
ok maybe this is clearer....
[40/(2+i)]-[35/(2-i)]
form the common denominator (2+i)(2-i) and
[40(2-i) -35(2+i)]/((2+i)(2-i))
= (80 - 40i - 70 - 35i)/(4-i^2)
=(10-75i)/5
= 2 - 15i
To solve this problem, you need to perform complex number division. Here are the steps to solve it:
Step 1: Multiply the complex conjugate of the denominator.
In this case, the complex conjugate of 2+i is 2-i, and the complex conjugate of 2-i is 2+i.
So, we need to multiply both the numerator and denominator by 2-i:
40 * (2-i) 35 * (2-i)
------------ --------------
(2+i) * (2-i) (2+i) * (2-i)
Step 2: Simplify each expression.
For the numerator:
40 * (2-i) = 80 - 40i
For the denominator:
(2+i) * (2-i) = 4 - 2i + 2i - (i^2)
= 4 - (i^2) (by applying the formula a^2 - b^2 = (a + b)(a - b))
= 4 - (-1) (since i^2 = -1)
= 4 + 1
= 5
So, the denominator simplifies to 5.
Step 3: Divide the simplified numerator by the simplified denominator.
(80 - 40i) / 5 = 80/5 - (40i)/5
= 16 - 8i
The solution to the division problem is 16 - 8i.