We are given that -i added to the product of (a+bi) and (2+3i) equals the product of (-11+5i) and (1-i)
In equation form we can write this as:
(a + bi)(2 + 3i) - i = (-11 + 5i)(1 - i)
2a + 3ai + 2bi + 3bi² - i = -11 +11i + 5i - 5i²
2a + i(3a + 2b - 1) - 3b = -11 + 16i + 5
2a - 3b + i(3a +2b -1) = -6 + 16i
Comparing the two sides, we can write:
2a - 3b = -6 (1)
and
3a + 2b - 1 = 16 (2)
Multiplying equation 1 by two and equation 2 by three and adding the results, we get:
2(2a - 3b) + 3(3a - 2b - 1)=2(-6) +3(16)
4a - 6b + 9a - 6b - 3 = -12 + 48
13a = 39
⇒ a= 3
Using the value of a in equation 1, we get:
2(3) - 3b = -6
-3b = -12
b = 4
Thus, the values of a and b satisfying the given condition are a = 3 and b = 4
