מתוך ויקיספר, אוסף הספרים והמדריכים החופשי
z
2
−
8
z
+
17
=
0
{\displaystyle z^{2}-8z+17=0}
z
2
−
2
z
+
26
=
0
{\displaystyle z^{2}-2z+26=0}
2
z
2
−
8
z
+
10
=
0
{\displaystyle 2z^{2}-8z+10=0}
z
2
+
(
1
−
i
)
z
−
i
=
0
{\displaystyle z^{2}+(1-i)z-i=0}
z
2
−
(
3
+
2
i
)
z
+
1
+
3
i
=
0
{\displaystyle z^{2}-(3+2i)z+1+3i=0}
z
2
−
i
z
−
9
+
3
i
=
0
{\displaystyle z^{2}-iz-9+3i=0}
z
2
+
i
z
+
6
=
0
{\displaystyle z^{2}+iz+6=0}
z
2
+
(
1
−
2
i
)
z
−
7
−
i
=
0
{\displaystyle z^{2}+(1-2i)z-7-i=0}
בנה משוואה ריבועית מתוקנת (כלומר, שמתקיים בה
a
=
1
{\displaystyle a=1}
z
1
=
2
i
,
z
2
=
i
−
1
{\displaystyle z_{1}=2i,z_{2}=i-1}
בנה משוואה ריבועית מתוקנת ששורשיה הם
z
1
=
3
+
2
i
,
z
2
=
i
−
4
{\displaystyle z_{1}=3+2i,z_{2}=i-4}
z
1
,
z
2
{\displaystyle z_{1},z_{2}}
13
z
2
−
8
z
+
2
=
0
{\displaystyle 13z^{2}-8z+2=0}
z
1
+
z
2
z
1
⋅
z
2
{\displaystyle {\sqrt {\frac {z_{1}+z_{2}}{z_{1}\cdot z_{2}}}}}
z
1
,
z
2
{\displaystyle z_{1},z_{2}}
z
2
−
(
3
+
2
i
)
z
+
2
i
=
0
{\displaystyle z^{2}-(3+2i)z+2i=0}
z
1
+
z
2
,
z
1
⋅
z
2
{\displaystyle z_{1}+z_{2},z_{1}\cdot z_{2}}
z
1
,
z
2
{\displaystyle z_{1},z_{2}}
z
2
−
(
4
+
2
i
)
z
+
2
=
0
{\displaystyle z^{2}-(4+2i)z+2=0}
1
z
1
,
1
z
2
{\displaystyle {\frac {1}{z_{1}}},{\frac {1}{z_{2}}}}
הוכח כי אם
z
1
,
z
2
{\displaystyle z_{1},z_{2}}
z
1
+
z
2
,
z
1
⋅
z
2
{\displaystyle z_{1}+z_{2},z_{1}\cdot z_{2}}
z
1
=
z
2
¯
{\displaystyle z_{1}={\overline {z_{2}}}}
z
1
,
2
=
4
±
i
{\displaystyle z_{1,2}=4\pm i}
z
1
,
2
=
1
±
5
i
{\displaystyle z_{1,2}=1\pm 5i}
z
1
,
2
=
2
±
i
{\displaystyle z_{1,2}=2\pm i}
z
1
=
−
1
,
z
2
=
i
{\displaystyle z_{1}=-1,z_{2}=i}
z
1
=
1
+
i
,
z
2
=
2
+
i
{\displaystyle z_{1}=1+i,z_{2}=2+i}
z
1
=
3
,
z
2
=
−
3
+
i
{\displaystyle z_{1}=3,z_{2}=-3+i}
z
1
=
2
i
,
z
2
=
−
3
i
{\displaystyle z_{1}=2i,z_{2}=-3i}
z
1
=
2
+
i
,
z
2
=
−
3
+
i
{\displaystyle z_{1}=2+i,z_{2}=-3+i}
z
2
+
(
1
−
3
i
)
z
−
2
(
1
+
i
)
=
0
{\displaystyle z^{2}+(1-3i)z-2(1+i)=0}
z
2
+
(
1
−
3
i
)
z
−
(
14
+
5
i
)
=
0
{\displaystyle z^{2}+(1-3i)z-(14+5i)=0}
2
{\displaystyle 2}
z
2
−
(
3
+
4
i
)
z
+
(
6
i
−
4
)
=
0
{\displaystyle z^{2}-(3+4i)z+(6i-4)=0}
2
z
2
−
(
4
+
2
i
)
z
+
1
=
0
{\displaystyle 2z^{2}-(4+2i)z+1=0}
נסמן
z
1
=
a
+
b
i
,
z
2
=
c
+
d
i
{\displaystyle z_{1}=a+bi,z_{2}=c+di}
z
2
{\displaystyle z_{2}}
d
≠
0
{\displaystyle d\neq 0}
z
1
+
z
2
=
(
a
+
c
)
+
(
b
+
d
)
i
{\displaystyle z_{1}+z_{2}=(a+c)+(b+d)i}
b
+
d
=
0
{\displaystyle b+d=0}
b
=
−
d
{\displaystyle b=-d}
כעת,
z
1
⋅
z
2
=
(
a
c
−
b
d
)
+
(
a
d
+
b
c
)
i
{\displaystyle z_{1}\cdot z_{2}=(ac-bd)+(ad+bc)i}
a
d
+
b
c
=
0
{\displaystyle ad+bc=0}
b
{\displaystyle b}
a
d
−
c
d
=
0
{\displaystyle ad-cd=0}
קיבלנו כי
a
d
=
c
d
{\displaystyle ad=cd}
z
2
{\displaystyle z_{2}}
d
≠
0
{\displaystyle d\neq 0}
a
=
c
{\displaystyle a=c}
כלומר, קיבלנו כי
z
1
=
a
+
b
i
{\displaystyle z_{1}=a+bi}
z
2
=
c
+
d
i
=
a
−
b
i
=
z
1
¯
{\displaystyle z_{2}=c+di=a-bi={\overline {z_{1}}}}
אם יש לנו משוואה ריבועיות שכל מקדמיה ממשיים ויש לה שורש לא ממשי
z
1
{\displaystyle z_{1}}
z
1
+
z
2
,
z
1
⋅
z
2
{\displaystyle z_{1}+z_{2},z_{1}\cdot z_{2}}
על פי מה שהראינו קודם, אם
z
1
+
z
2
,
z
1
⋅
z
2
{\displaystyle z_{1}+z_{2},z_{1}\cdot z_{2}}
z
1
=
z
2
¯
{\displaystyle z_{1}={\overline {z_{2}}}}
