Question

решение комплексных чисел​

Answer 1

$1)\ \ z_1=3+5i\ \ ,\ \ z_2=4-i\\\\z_1+z_2=(3+4)+(5i-i)=7+4i\\\\z_1-z_2=(3-4)+(5i-(-i))=-1+6i\\\\z_1\cdot z_2=(3+5i)(4-i)=12-3i+20i-5i^2=17+17i\ \ \ (i^2=-1)\\\\\dfrac{z_1}{z_2}=\dfrac{(3+5i)(4+i)}{(4-i)(4+i)}=\dfrac{12+3i+20i+5i^2}{16-i^2}=\dfrac{7+23i}{16+1}=\dfrac{7}{17}+\dfrac{23}{17}\, i$

$z_1^3=(3+5i)^3=27+135i+225i^2+125i^3=27+135i-225-125i=\\\\=-198+10i\\\\z_2^2=(4-i)^2=16-8i+i^2=15-8i$

$2)\ \ i^{16}-2i^3+6=(i^2)^8-2i^2\cdot i+6=(-1)^8+2i+6=1+2i+6=7+2i\\\\4\, i^7-i+i^2=4\, i^6\cdot i-i-1=4\, (i^2)^3\cdot i-i-1=-4\cdt i-i-1=-5i-1\\\\(i-1)^2+i^{88}=i^2-2i+1+(i^2)^{44}=-1-2i+1+(-1)^{44}=-2i+1=1-2i$

$3)\ \ z_1=-8+i\ \ ,\ \ \overline {z_1}=-8-i\\\\z_2=4-3i\ \ ,\ \ \overline {z_2}=4+3i\\\\z_3=5i\ \ ,\ \ \ \overline {z_3}=-5i$