Izibonelo Zemibuzo Nezingxoxo Ngemisebenzi Yezinombolo Eziyinkimbinkimbi
Izinombolo eziyinkimbinkimbi ziyisandiso somqondo wezinombolo zangempela ukuze kufakwe izinombolo ezicatshangelwayo. Uhlobo olujwayelekile lwenombolo eyinkimbinkimbi luyi-a + bi, lapho u-a no-b beyizinombolo zangempela, kanti u-i uyiyunithi ecatshangelwayo enesici i² = -1. Imisebenzi enombolweni eyinkimbinkimbi ihlanganisa ukuhlanganisa, ukususa, ukuphindaphinda, nokuhlukanisa. Lesi sihloko sizohlinzeka ngezibonelo eziningana zezinkinga kanye nezingxoxo zemisebenzi ehlukahlukene enombolweni eyinkimbinkimbi.
Ukwengeza nokususa izinombolo eziyinkimbinkimbi
Isibonelo Umbuzo 1
Engeza izinombolo eziyinkimbinkimbi ezilandelayo: (3 + 4i) kanye no-(1 + 2i).
Ingxoxo:
Ukwengeza izinombolo eziyinkimbinkimbi kwenziwa ngokungeza izingxenye zazo zangempela nezicatshangwayo ngokwahlukana.
\[ (3 + 4i) + (1 + 2i) = (3 + 1) + (4i + 2i) = 4 + 6i \]
Ngakho-ke, umphumela wokwengeza u-(3 + 4i) kanye no-(1 + 2i) ngu-4 + 6i.
Isibonelo Umbuzo 2
Susa inombolo eyinkimbinkimbi (2 + 5i) ku-(6 + 3i).
Ingxoxo:
Ukususa izinombolo eziyinkimbinkimbi kwenziwa ngokukhipha ingxenye yangempela kanye nengxenye ecatshangelwayo ngokwehlukana.
\[ (6 + 3i) – (2 + 5i) = (6 – 2) + (3i – 5i) = 4 – 2i \]
Ngakho-ke, umphumela wokususa u-(2 + 5i) ku-(6 + 3i) u-4 – 2i.
Ukuphindaphinda Kwezinombolo Eziyinkimbinkimbi
Isibonelo Umbuzo 3
Phindaphinda izinombolo eziyinkimbinkimbi ezilandelayo: (2 + 3i) kanye no (4 + i).
Ingxoxo:
Ukuphindaphinda kwezinombolo eziyinkimbinkimbi kwenziwa kusetshenziswa ukusatshalaliswa noma amalungiselelo asemthethweni, okufana nokuphindaphinda ama-binomial amabili ku-algebra evamile.
\[
(2 + 3i) \cdot (4 + i) = 2 \cdot 4 + 2 \cdot i + 3i \cdot 4 + 3i \cdot i
\]
Bese sibala ngokuningiliziwe:
\[
= 8 + 2i + 12i + 3i^2
\]
Kusukela \( i^2 = -1 \):
\[
= 8 + 14i + 3(-1)
\]
\[
= 8 + 14i – 3
\]
\[
= 5 + 14i
\]
Ngakho-ke, umphumela wokuphindaphinda (2 + 3i) kanye no-(4 + i) ngu-5 + 14i.
Ukuhlukaniswa Kwezinombolo Eziyinkimbinkimbi
Isibonelo Umbuzo 4
Hlukanisa inombolo eyinkimbinkimbi elandelayo: (5 + 6i) ngo (2 + i).
Ingxoxo:
Ukuhlukaniswa kwezinombolo eziyinkimbinkimbi kusetshenziswa i-conjugate ye-denominator. I-conjugate ye-\(2 + i\) ingu-\(2 – i\).
Siphindaphinda inombolo kanye nenani eliyisiqalo nge-conjugate yamanani aphansi:
\[
\frac{5 + 6i}{2 + i} \cdot \frac{2 – i}{2 – i}
\]
Manje sibala inombolo kanye nenani eliphakathi ngokwehlukana:
\[
= \frac{(5 + 6i) \cdot (2 – i)}{(2 + i) \cdot (2 – i)}
\]
Ukuphindaphinda kwama-denominator:
\[
(2 + i) \cdot (2 – i) = 2^2 – i^2 = 4 – (-1) = 4 + 1 = 5
\]
Ukuphindaphinda kwezinombolo:
\[
(5 + 6i) \cdot (2 – i) = 5 \cdot 2 + 5 \cdot (-i) + 6i \cdot 2 + 6i \cdot (-i)
= 10 – 5i + 12i – 6i^2
= 10 + 7i – 6(-1)
= 10 + 7i + 6
= 16 + 7i
\]
Ngakho-ke, isigaba sithi:
\[
= \frac{16 + 7i}{5} = \frac{16}{5} + \frac{7i}{5} = 3.2 + 1.4i
\]
Ngakho-ke umphumela wokuhlukanisa (5 + 6i) ngo-(2 + i) ngu-3.2 + 1.4i.
Ingxoxo Eyengeziwe: I-Modulus kanye ne-Conjugate yezinombolo Eziyinkimbinkimbi
Isibonelo Umbuzo 5
Thola i-modulus kanye ne-conjugate yenombolo eyinkimbinkimbi \(z = 3 + 4i\).
Ingxoxo:
I-modulus yenombolo eyinkimbinkimbi \(z = a + bi\) ithi:
\[
|z| = \sqrt{a^2 + b^2}
\]
Ku-\(z = 3 + 4i\):
\[
|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
I-conjugate yenombolo eyinkimbinkimbi \(z = a + bi\) ngu \(z^ = a – bi\).
Ku-\(z = 3 + 4i\):
\[
z^ = 3 – 4i
\]
Ngakho-ke i-modulus ka-\(3 + 4i\) ingu-5, kanti i-conjugate yayo ingu-\(3 - 4i\).
Isiphetho
Izinombolo eziyinkimbinkimbi zidlala indima ebalulekile emikhakheni eyahlukahlukene yezibalo kanye nezinhlelo zokusebenza zobuchwepheshe. Ukuqonda imisebenzi eyisisekelo enombolweni eziyinkimbinkimbi, njengokuhlanganisa, ukususa, ukuphindaphinda, kanye nokuhlukanisa, kubalulekile ekusebenziseni le mibono ekuxazululeni izinkinga eziyinkimbinkimbi kakhulu. Ukuzijwayeza izinhlobo ezahlukene zezinkinga, njengalezo ezichazwe ngenhla, kuzosiza ukuqinisa ukuqonda kwakho kanye namakhono okusebenza ngezinombolo eziyinkimbinkimbi.
