What is the answer to (3z+i)/((2+i)z+1)=i ?

The answer to (3z+i)/((2+i)z+1)=i is z=0

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(3z+i)/((2+i)z+1)=i

\frac{3 z + i}{( 2 + i ) z + 1} = i

z = 0

Solution steps

\frac{3 z + i}{( 2 + i ) z + 1} = i

Substitute

z = x + y i

\frac{3 ( x + y i ) + i}{( 2 + i ) ( x + y i ) + 1} = i

Multiply both sides by

(2 + i) (x + y i) + 1

\frac{3 ( x + y i ) + i}{( 2 + i ) ( x + y i ) + 1} ((2 + i) (x + y i) + 1) = i ((2 + i) (x + y i) + 1)

Expand

3 x + i (1 + 3 y) = (- 2 y - x) + i (1 - y + 2 x)

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

[3 x = - 2 y - x 1 + 3 y = 1 - y + 2 x]

[3 x = - 2 y - x 1 + 3 y = 1 - y + 2 x] : x = 0, y = 0

Substitute back

z = x + y i

z = 0

5 x^{2} - 11 i x - 2 = 0

\frac{2 ( x + i y )}{3 - ( x - i y )} = i

(3 x - y) + (x + y) i = 12

z^{2} - (1 + 3 i) z + (2 i - 2) = 0

\frac{1 + ai}{4 + bi} = 1 + 4 i

What is the answer to (3z+i)/((2+i)z+1)=i ?
The answer to (3z+i)/((2+i)z+1)=i is z=0