What is the answer to 2a-bi= 1/(sqrt(3)-4i) ?

The answer to 2a-bi= 1/(sqrt(3)-4i) is a=(sqrt(3))/(38),b=-4/19

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2a-bi= 1/(sqrt(3)-4i)

2 a - bi = \frac{1}{3 - 4 i}

a = \frac{3}{38}, b = - \frac{4}{19}

Solution steps

2 a - bi = \frac{1}{3 - 4 i}

Expand

\frac{1}{3 - 4 i} : \frac{3}{19} + i \frac{4}{19}

2 a - bi = \frac{3}{19} + i \frac{4}{19}

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

[2 a = \frac{3}{19} - b = \frac{4}{19}]

[2 a = \frac{3}{19} - b = \frac{4}{19}] : a = \frac{3}{38}, b = - \frac{4}{19}

a = \frac{3}{38}, b = - \frac{4}{19}

x^{2} - (5 - i) x + (18 + i) = 0

2 x - 3 y i = 14 + 12 i

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

i = - 5 x^{2} + 10 x + 10

M = C (1 + i t)

What is the answer to 2a-bi= 1/(sqrt(3)-4i) ?
The answer to 2a-bi= 1/(sqrt(3)-4i) is a=(sqrt(3))/(38),b=-4/19