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

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

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

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

a = - 3 b, b = b

Solution steps

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

Expand

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

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

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

[3 a - b = 3 a - b a + 3 b = - a - 3 b] : a = - 3 b, b = b

a = - 3 b, b = b

z^{2} - (3 + 4 i) z + 7 i - 1 = 0

- 3 \cdot 2 + 30 x - 75 s i x = 0

w^{2} = - 40 i

- \frac{1}{z - i} - 1 - i - z = 0

+ 3 i z^{2} = - 7 + 8 i, z^{1} = 5

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