1.If α\alphaα and β\betaβ are the roots of the equation 2z2−3z−2i=02z^2 - 3z - 2i = 02z2−3z−2i=0, where i=−1i = \sqrt{-1}i=−1, then Re (α19+β19+α11+β11α15+β15)\mathrm{Re}\!\left(\dfrac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}}\right)Re(α15+β15α19+β19+α11+β11) is equal to:a.441441441b.398398398c.312312312d.409409409Login to continueOnly logged in users canattempt or see the solution.