Given,

$Z_L=60-j80 Ω \\ Z_0=50 Ω \\ d_2=λ/8$

Since $d_1$ is not given, assume $d_1=0$

$\mathcal{Z}_L=\frac{Z_L}{Z_0} =\frac{60-j80}{50}=1.2-j1.6 Ω \\ y_L=\frac{1}{\mathcal{Z}_L} =0.3+j0.4$

Since $d_1=0 ∴y_L=y_{d1}$

$y_{d1}$ in terms of $λ→0.0645 λ$

$y_{d1}=0.3+j0.4$

For spacing circle

$2βdl=2×\frac{2π}{λ}×\frac{λ}{8}=90° (or-90°)$

From smith chart 3,

$y_{11}=0.3-j0.3 \\ y_{22}=0.3-j1.7 \\ y_{s1}=y_{11}-y_{d1} \\ y_{s1}=-j0.7 \\ y_{s1}^`=y_{11}^`-y_{d1} \\ y_{s1}^`=-j2.1$

For open circuit stub, we obtain from smith chart 4,

$l_1=0.403 λ \\ l_1^`=0.321 λ$

From smith chart 1

$y_{d2}=1+j1.5 \\ y_{d2}^`=1-j3.2$

In the direction of VSWR circle,

$y_{s2}=1-y_{d2} \\ y_{s2}=-j1.5 \\ y_{s2}^`=1-y_{d2}^` \\ y_{s2}^`=j3.2$

For open circuit stub, we obtain from smith chart 4,

$l_2=0.344 λ \\ l_2^`=0.201 λ$

Results,

$l_1=0.403 λ \\ l_1^`=0.321 λ \\ l_2=0.344 λ \\ l_2^`=0.201 λ$

**Stub design-**