What is the answer to i(a+bi-5)=(1+3i)(a+bi) ?

The answer to i(a+bi-5)=(1+3i)(a+bi) is a=-2,b=-1

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i(a+bi-5)=(1+3i)(a+bi)

i (a + bi - 5) = (1 + 3 i) (a + bi)

a = - 2, b = - 1

Solution steps

i (a + bi - 5) = (1 + 3 i) (a + bi)

Expand

- b + i (a - 5) = (a - 3 b) + i (b + 3 a)

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

[- b = a - 3 b a - 5 = b + 3 a]

[- b = a - 3 b a - 5 = b + 3 a] : a = - 2, b = - 1

a = - 2, b = - 1

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

\frac{4 + ai}{2 - i} = b - 4 i

r (t) = t^{3} i + 2 t^{2} j

v = v i t + \frac{1}{2} g t^{2}

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

What is the answer to i(a+bi-5)=(1+3i)(a+bi) ?
The answer to i(a+bi-5)=(1+3i)(a+bi) is a=-2,b=-1