To find a and b in 2+2a+3i+3ib = 6a+9i

Solution:

Simplify left sides such that they are grouped in real and imaginary numbers.

LHS : (2+2a) + (3+3b)i

RHS : 6a+9i.

Equate the real number on left side with real number on right side: 2+2a = 6a. Solve for a.

2 = 6a-2a

2 = 4a

2/4 = a . Or a = 1/2.

Equate the imaginary number on the left with the imaginary number on the right side:

(3+3b)i = 9i. Or

3+3b = 9. Solve for b:

3b = 9-3 = 2

3b = 6. Divide by 2.

b = 2.

To calculate a and b, we'll group, both sides, the real parts and the imaginary parts.

First, we'll gather all the terms in a and b, to the left side and the rest of terms, we'll move to the right side.

2a + 3ib - 6a = -2 - 3i + 9i

We'll combine like terms:

-4a + 3ib = -2 + 6i

To the left side, the real part is -4a and the imaginary part is 3b.

We'll put the real part from the left side equal to the real part from the right side:

-4a = -2

We'll divide by -4:

a = -2/-4

**a = 1/2**

We'll put the imaginary part from the left side equal to the imaginary part from the right side:

3b = 6

We'll divide by 3:

**b = 2**