What is the answer to (3-a+2b)/5-3i+a+bi=4-i ?

The answer to (3-a+2b)/5-3i+a+bi=4-i is a= 13/4 ,b=2

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(3-a+2b)/5-3i+a+bi=4-i

\frac{3 - a + 2 b}{5} - 3 i + a + bi = 4 - i

a = \frac{13}{4}, b = 2

Solution steps

\frac{3 - a + 2 b}{5} - 3 i + a + bi = 4 - i

Multiply both sides by

5

\frac{3 - a + 2 b}{5} \cdot 5 - 3 i 5 + a \cdot 5 + bi 5 = 4 \cdot 5 - i 5

Simplify

(4 a + 2 b + 3) + i (- 15 + 5 b) = 20 - 5 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[4 a + 2 b + 3 = 20 - 15 + 5 b = - 5]

[4 a + 2 b + 3 = 20 - 15 + 5 b = - 5] : a = \frac{13}{4}, b = 2

a = \frac{13}{4}, b = 2

(7 - 7 i) - 8 = x - (2 + y i)

i (x) = \frac{3}{x + 3}

\frac{1 - i}{1 + z \cdot i} = 2 - 3 i

(9 x^{2} - 9 y^{2}) + 18 x y i = 32 - 24

(2 + 2 i) (x + i y) = (2 + i)^{2}

What is the answer to (3-a+2b)/5-3i+a+bi=4-i ?
The answer to (3-a+2b)/5-3i+a+bi=4-i is a= 13/4 ,b=2