What is the answer to +6iz^2=3+5i,z^1=2 ?

The answer to +6iz^2=3+5i,z^1=2 is

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+6iz^2=3+5i,z^1=2

+ 6 i z^{2} = 3 + 5 i, z^{1} = 2

z = \frac{5 + 34}{2 3} - \frac{3}{2 5 + 34} i, z = - \frac{5 + 34}{2 3} + \frac{3}{2 5 + 34} i

Solution steps

6 i z^{2} = 3 + 5 i

Substitute

z = x + y i

6 i (x + y i)^{2} = 3 + 5 i

Expand

6 i (x + y i)^{2} : - 12 x y + 6 i (x^{2} - y^{2})

- 12 x y + 6 i (x^{2} - y^{2}) = 3 + 5 i

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

[- 12 x y = 3 6 x^{2} - 6 y^{2} = 5]

[- 12 x y = 3 6 x^{2} - 6 y^{2} = 5] : x = \frac{5 + 34}{2 3}, x = - \frac{5 + 34}{2 3}, y = - \frac{3}{2 5 + 34} y = \frac{3}{2 5 + 34}

Substitute back

z = x + y i

z = \frac{5 + 34}{2 3} - \frac{3}{2 5 + 34} i, z = - \frac{5 + 34}{2 3} + \frac{3}{2 5 + 34} i

(2 + 3 i) (a + bi) = 13

22 (x + y i) + (28 + 4 i) = 72 - 62 i

a x^{2} - b x + 9 + 3 i = 0

2 i (a + bi) + 6 = 9 i + (a + bi)

(x + i y)^{2} = 3 + 2 i