What is the answer to (x-(3-4i))(x-(3+4i))=0 ?

The answer to (x-(3-4i))(x-(3+4i))=0 is x=3+4i,x=3-4i

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(x-(3-4i))(x-(3+4i))=0

(x - (3 - 4 i)) (x - (3 + 4 i)) = 0

x = 3 + 4 i, x = 3 - 4 i

Solution steps

(x - (3 - 4 i)) (x - (3 + 4 i)) = 0

Substitute

x = a + bi

(a + bi - (3 - 4 i)) (a + bi - (3 + 4 i)) = 0

Expand

(a + bi - (3 - 4 i)) (a + bi - (3 + 4 i)) : (a^{2} - 6 a + 25 - b^{2}) + i (- 6 b + 2 ab)

(a^{2} - 6 a + 25 - b^{2}) + i (- 6 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - 6 a + 25 - b^{2}) + i (- 6 b + 2 ab) = 0 + 0 i

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

[a^{2} - 6 a + 25 - b^{2} = 0 - 6 b + 2 ab = 0]

[a^{2} - 6 a + 25 - b^{2} = 0 - 6 b + 2 ab = 0] : (a = 3, a = 3, b = 4 b = - 4)

Substitute back

x = a + bi

x = 3 + 4 i, x = 3 - 4 i

(x + i) (3 - i y) = 1 + 13 i

(a - bi) + (b + ai) = 5 + i

z^{2} - 2 z = 3 i

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

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

What is the answer to (x-(3-4i))(x-(3+4i))=0 ?
The answer to (x-(3-4i))(x-(3+4i))=0 is x=3+4i,x=3-4i