Question

Z1=2-3i
Z2=4-i
Z1+Z2
Z1-Z2
Z1*Z2
Z1/Z2
Z1²
Z1³
Z2²
Z2³

Answer 1

$z_1=2-3i\\z_2=4-i$

$z_1+z_2=2-3i+4-i=6-4i\\ z_1-z_2=2-3i-(4-i)=2-3i-4+i=-2-2i\\z_1 \cdot z_2=(2-3i)(4-i)=8-12i-2i+3i^2=8-14i-3=5-14i\\\dfrac{z_1}{z_2}=\dfrac{2-3i}{4-i}=\dfrac{(2-3i)(4+i)}{(4-i)(4+i)}=\dfrac{8-12i+2i+3}{16-i^2}=\dfrac{11-10i}{17}=\dfrac{11}{17}-\dfrac{10}{17} \,i\\z_1^2=(2-3i)^2=4-12i+9i^2=4-12i-9=-5-12i\\z_1^3=z_1^2 \cdot z_1=(-5-12i)(2-3i)=-(5+12i)(2-3i)=\\-(10+24i-15i+36)=-(46+9i)=-46-9i\\z_2^2=(4-i)^2=16-8i+i^2=16-8i-1=15-8i\\z_2^3=z_2^2 \cdot z_2=(15-8i)(4-i)=60-32i-15i-8=52-47i$