What is the answer to (2a+1)(3+2i)+8=bi ?

The answer to (2a+1)(3+2i)+8=bi is a=-11/6 ,b=-16/3

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(2a+1)(3+2i)+8=bi

(2 a + 1) (3 + 2 i) + 8 = bi

a = - \frac{11}{6}, b = - \frac{16}{3}

Solution steps

(2 a + 1) (3 + 2 i) + 8 = bi

Rewrite

(2 a + 1) (3 + 2 i) + 8

in standard complex form:

(6 a + 11) + (4 a + 2) i

(6 a + 11) + (4 a + 2) i = bi

Rewrite

bi

in standard complex form:

0 + bi

(6 a + 11) + (4 a + 2) i = 0 + bi

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

[6 a + 11 = 0 4 a + 2 = b]

[6 a + 11 = 0 4 a + 2 = b] : a = - \frac{11}{6}, b = - \frac{16}{3}

a = - \frac{11}{6}, b = - \frac{16}{3}

(1 + i) (x + i y) = 2 - 5 i

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

t, a = v f - \frac{v i}{t}

1.2 x + 3 + y i = 7.2 - 4.3 i

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

What is the answer to (2a+1)(3+2i)+8=bi ?
The answer to (2a+1)(3+2i)+8=bi is a=-11/6 ,b=-16/3