If the set Re(z-z+zz2-3z+5z):z C,Re(z)=3 is equal to the interval ( , ], then 24( - ) is e…

Question

If the set ${Re(\frac{z-\vec{z}+z\vec{z}}{2-3z+5\vec{z}}):z\in C,Re(z)=3}$ is equal to

the interval $(\alpha ,\beta ]$, then $24(\beta -\alpha )$ is equal to :

Accepted Answer

Let ${z}_{1}=(\frac{z-\vec{z}+z\vec{z}}{2-3z+5\vec{z}})$

Let $z=3+iy$

$\vec{z}=3-iy$

$\begin{gathered} {z}_{1}=\frac{2iy+(9+{y}^{2})}{2-3(3+iy)+5(3-iy)} \ =\frac{9+{y}^{2}+i(2y)}{8-8iy} \ =\frac{(9+{y}^{2})+i(2y)}{8(1-iy)} \ Re({z}_{1})=\frac{(9+{y}^{2})-2{y}^{2}}{8(1+{y}^{2})} \end{gathered}$

$\begin{gathered} =\frac{9-{y}^{2}}{8(1+{y}^{2})} \ =\frac{1}{8}[\frac{10-(1+{y}^{2})}{(1+{y}^{2})}] \ =\frac{1}{8}[\frac{10}{1+{y}^{2}}-1] \ 1+{y}^{2}\in [1,\infty] \ \frac{1}{1+{y}^{2}}\in (0,1] \ \frac{10}{1+{y}^{2}}\in (0,10] \ \frac{10}{1+{y}^{2}}-1\in (-1,9] \ Re({z}_{1})\in (\frac{-1}{8},\frac{9}{8}] \ \alpha =\frac{-1}{8},\beta =\frac{9}{8} \ 24(\beta -\alpha )=24(\frac{9}{8}+\frac{1}{8})=30 \end{gathered}$

Solutions · 1