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(a+b*i)*(a+b*i)-(1+i)*(a+b*i)+i=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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(a + b*I)*(a + b*I) - (1 + I)*(a + b*I) + I = 0
$$\left(\left(a + i b\right) \left(a + i b\right) - \left(1 + i\right) \left(a + i b\right)\right) + i = 0$$
Detail solution
Expand the expression in the equation
$$\left(\left(a + i b\right) \left(a + i b\right) - \left(1 + i\right) \left(a + i b\right)\right) + i = 0$$
We get the quadratic equation
$$a^{2} + 2 i a b - a - i a - b^{2} + b - i b + i = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$b_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$b_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 2 i a + 1 - i$$
$$c = a^{2} - a - i a + i$$
, then
The equation has two roots.
or
$$b_{1} = i a - \frac{\sqrt{4 a^{2} - 4 a - 4 i a + \left(2 i a + 1 - i\right)^{2} + 4 i}}{2} + \frac{1}{2} - \frac{i}{2}$$
$$b_{2} = i a + \frac{\sqrt{4 a^{2} - 4 a - 4 i a + \left(2 i a + 1 - i\right)^{2} + 4 i}}{2} + \frac{1}{2} - \frac{i}{2}$$
$$\left(\left(a + i b\right) \left(a + i b\right) - \left(1 + i\right) \left(a + i b\right)\right) + i = 0$$
We get the quadratic equation
$$a^{2} + 2 i a b - a - i a - b^{2} + b - i b + i = 0$$
This equation is of the form
a*b^2 + b*b + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$b_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$b_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 2 i a + 1 - i$$
$$c = a^{2} - a - i a + i$$
, then
D = b^2 - 4 * a * c =
(1 - i + 2*i*a)^2 - 4 * (-1) * (i + a^2 - a - i*a) = (1 - i + 2*i*a)^2 - 4*a + 4*i + 4*a^2 - 4*i*a
The equation has two roots.
b1 = (-b + sqrt(D)) / (2*a)
b2 = (-b - sqrt(D)) / (2*a)
or
$$b_{1} = i a - \frac{\sqrt{4 a^{2} - 4 a - 4 i a + \left(2 i a + 1 - i\right)^{2} + 4 i}}{2} + \frac{1}{2} - \frac{i}{2}$$
$$b_{2} = i a + \frac{\sqrt{4 a^{2} - 4 a - 4 i a + \left(2 i a + 1 - i\right)^{2} + 4 i}}{2} + \frac{1}{2} - \frac{i}{2}$$
The graph
Sum and product of roots
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sum
-im(a) + I*(-1 + re(a)) + 1 - im(a) + I*re(a)
$$\left(i \left(\operatorname{re}{\left(a\right)} - 1\right) - \operatorname{im}{\left(a\right)}\right) + \left(i \operatorname{re}{\left(a\right)} - \operatorname{im}{\left(a\right)} + 1\right)$$
=
1 - 2*im(a) + I*(-1 + re(a)) + I*re(a)
$$i \left(\operatorname{re}{\left(a\right)} - 1\right) + i \operatorname{re}{\left(a\right)} - 2 \operatorname{im}{\left(a\right)} + 1$$
product
(-im(a) + I*(-1 + re(a)))*(1 - im(a) + I*re(a))
$$\left(i \left(\operatorname{re}{\left(a\right)} - 1\right) - \operatorname{im}{\left(a\right)}\right) \left(i \operatorname{re}{\left(a\right)} - \operatorname{im}{\left(a\right)} + 1\right)$$
=
(-im(a) + I*(-1 + re(a)))*(1 - im(a) + I*re(a))
$$\left(i \left(\operatorname{re}{\left(a\right)} - 1\right) - \operatorname{im}{\left(a\right)}\right) \left(i \operatorname{re}{\left(a\right)} - \operatorname{im}{\left(a\right)} + 1\right)$$
(-im(a) + i*(-1 + re(a)))*(1 - im(a) + i*re(a))
Rapid solution
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b1 = -im(a) + I*(-1 + re(a))
$$b_{1} = i \left(\operatorname{re}{\left(a\right)} - 1\right) - \operatorname{im}{\left(a\right)}$$
b2 = 1 - im(a) + I*re(a)
$$b_{2} = i \operatorname{re}{\left(a\right)} - \operatorname{im}{\left(a\right)} + 1$$
b2 = i*re(a) - im(a) + 1