What is the answer to a^2+b^2+1+12i=6(a+bi) ?

The answer to a^2+b^2+1+12i=6(a+bi) is

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a^2+b^2+1+12i=6(a+bi)

a^{2} + b^{2} + 1 + 12 i = 6 (a + bi)

(a = 5, a = 1, b = 2 b = 2)

Solution steps

a^{2} + b^{2} + 1 + 12 i = 6 (a + bi)

Rewrite

6 (a + bi)

in standard complex form:

6 a + 6 bi

a^{2} + b^{2} + 1 + 12 i = 6 a + 6 bi

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

[a^{2} + b^{2} + 1 = 6 a 12 = 6 b]

[a^{2} + b^{2} + 1 = 6 a 12 = 6 b] : (a = 5, a = 1, b = 2 b = 2)

(a = 5, a = 1, b = 2 b = 2)

Z^{2} - (2 + 2 i) Z - 4 + 2 i = 0

(a + bi)^{2} = 15 + 8 i

i = - 0.3 q^{2} + 30 q

x, s i x = x (x + 1) - x (x - 1)

(z - (2 + i)) (z - (2 - i)) = 0

What is the answer to a^2+b^2+1+12i=6(a+bi) ?
The answer to a^2+b^2+1+12i=6(a+bi) is