z=y4sin5x+2x−2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
z1 = -2 + 2*re(x) + I*(2*im(x) + im(y4*sin(5*x))) + re(y4*sin(5*x))
$$z_{1} = i \left(2 \operatorname{im}{\left(x\right)} + \operatorname{im}{\left(y_{4} \sin{\left(5 x \right)}\right)}\right) + 2 \operatorname{re}{\left(x\right)} + \operatorname{re}{\left(y_{4} \sin{\left(5 x \right)}\right)} - 2$$
z1 = i*(2*im(x) + im(y4*sin(5*x))) + 2*re(x) + re(y4*sin(5*x)) - 2
Sum and product of roots
[src]
-2 + 2*re(x) + I*(2*im(x) + im(y4*sin(5*x))) + re(y4*sin(5*x))
$$i \left(2 \operatorname{im}{\left(x\right)} + \operatorname{im}{\left(y_{4} \sin{\left(5 x \right)}\right)}\right) + 2 \operatorname{re}{\left(x\right)} + \operatorname{re}{\left(y_{4} \sin{\left(5 x \right)}\right)} - 2$$
-2 + 2*re(x) + I*(2*im(x) + im(y4*sin(5*x))) + re(y4*sin(5*x))
$$i \left(2 \operatorname{im}{\left(x\right)} + \operatorname{im}{\left(y_{4} \sin{\left(5 x \right)}\right)}\right) + 2 \operatorname{re}{\left(x\right)} + \operatorname{re}{\left(y_{4} \sin{\left(5 x \right)}\right)} - 2$$
-2 + 2*re(x) + I*(2*im(x) + im(y4*sin(5*x))) + re(y4*sin(5*x))
$$i \left(2 \operatorname{im}{\left(x\right)} + \operatorname{im}{\left(y_{4} \sin{\left(5 x \right)}\right)}\right) + 2 \operatorname{re}{\left(x\right)} + \operatorname{re}{\left(y_{4} \sin{\left(5 x \right)}\right)} - 2$$
-2 + 2*re(x) + I*(2*im(x) + im(y4*sin(5*x))) + re(y4*sin(5*x))
$$i \left(2 \operatorname{im}{\left(x\right)} + \operatorname{im}{\left(y_{4} \sin{\left(5 x \right)}\right)}\right) + 2 \operatorname{re}{\left(x\right)} + \operatorname{re}{\left(y_{4} \sin{\left(5 x \right)}\right)} - 2$$
-2 + 2*re(x) + i*(2*im(x) + im(y4*sin(5*x))) + re(y4*sin(5*x))