Nā Laʻana o nā Nīnau a me nā Kūkākūkā e pili ana i nā Hana ma nā Helu Paʻakikī
ʻO nā helu paʻakikī kahi hoʻonui o ke kumumanaʻo o nā helu maoli e hoʻokomo i nā helu hoʻokalakupua. ʻO ke ʻano maʻamau o kahi helu paʻakikī ʻo a + bi, kahi ʻo a a me b he mau helu maoli, a ʻo i kahi ʻāpana hoʻokalakupua me ka waiwai i² = -1. ʻO nā hana ma nā helu paʻakikī e komo pū me ka hoʻohui, ka unuhi, ka hoʻonui, a me ka mahele. E hāʻawi kēia ʻatikala i kekahi mau pilikia hoʻohālike a me nā kūkākūkā no nā hana like ʻole ma nā helu paʻakikī.
Hoʻohui a me ka Hoʻemi ʻana o nā Helu Paʻakikī
Laʻana Nīnau 1
E hoʻohui i nā helu paʻakikī aʻe: (3 + 4i) a me (1 + 2i).
Kūkākūkā:
Hana ʻia ka hoʻohui ʻana i nā helu paʻakikī ma ka hoʻohui ʻana i ko lākou mau ʻāpana maoli a me nā ʻāpana hoʻokalakupua ma ke kaʻawale.
\[ (3 + 4i) + (1 + 2i) = (3 + 1) + (4i + 2i) = 4 + 6i \]
No laila, ʻo ka hopena o ka hoʻohui ʻana i (3 + 4i) a me (1 + 2i) he 4 + 6i.
Laʻana Nīnau 2
E unuhi i ka helu paʻakikī (2 + 5i) mai (6 + 3i).
Kūkākūkā:
Hana ʻia ka unuhi ʻana o nā helu paʻakikī ma ka unuhi ʻana i ka ʻāpana maoli a me ka ʻāpana hoʻokalakupua ma ke kaʻawale.
\[ (6 + 3i) – (2 + 5i) = (6 – 2) + (3i – 5i) = 4 – 2i \]
No laila, ʻo ka hopena o ka unuhi ʻana i (2 + 5i) mai (6 + 3i) he 4 - 2i.
Hoʻonui ʻia o nā helu paʻakikī
Laʻana Nīnau 3
E hoʻonui i nā helu paʻakikī aʻe: (2 + 3i) a me (4 + i).
Kūkākūkā:
Hana ʻia ka hoʻonui ʻana o nā helu paʻakikī me ka hoʻohana ʻana i nā hoʻolaha a i ʻole nā hoʻonohonoho kūhelu, e like me ka hoʻonui ʻana i ʻelua binomials i ka algebra maʻamau.
\[
(2 + 3i) \cdot (4 + i) = 2 \cdot 4 + 2 \cdot i + 3i \cdot 4 + 3i \cdot i
\]
A laila mākou e helu pono ai:
\[
= 8 + 2i + 12i + 3i^2
\]
ʻOiai \( i^2 = -1 \):
\[
= 8 + 14i + 3(-1)
\]
\[
= 8 + 14i – 3
\]
\[
= 5 + 14i
\]
No laila, ʻo ka hopena o ka hoʻonui ʻana iā (2 + 3i) a me (4 + i) he 5 + 14i.
Ka Māhele ʻana o nā Helu Paʻakikī
Laʻana Nīnau 4
E puʻunaue i ka helu paʻakikī aʻe: (5 + 6i) me (2 + i).
Kūkākūkā:
ʻO ka mahele ʻana o nā helu paʻakikī me ka hoʻohana ʻana i ka conjugate o ka denominator. ʻO ka conjugate o \(2 + i\) ʻo \(2 – i\).
Hoʻonui mākou i ka numerator a me ka denominator e ka conjugate o nā denominators:
\[
\frac{5 + 6i}{2 + i} \cdot \frac{2 – i}{2 – i}
\]
I kēia manawa ke helu nei mākou i ka numerator a me ka denominator ma ke kaʻawale:
\[
= \frac{(5 + 6i) \cdot (2 – i)}{(2 + i) \cdot (2 – i)}
\]
Hoʻonui ʻia o nā denominators:
\[
(2 + i) \cdot (2 – i) = 2^2 – i^2 = 4 – (-1) = 4 + 1 = 5
\]
Hoʻonui ʻia o nā helu:
\[
(5 + 6i) \cdot (2 – i) = 5 \cdot 2 + 5 \cdot (-i) + 6i \cdot 2 + 6i \cdot (-i)
= 10 – 5i + 12i – 6i^2
= 10 + 7i – 6(-1)
= 10 + 7i + 6
= 16 + 7i
\]
No laila, ʻo ka mahele ʻana:
\[
= \frac{16 + 7i}{5} = \frac{16}{5} + \frac{7i}{5} = 3.2 + 1.4i
\]
No laila, ʻo ka hopena o ka puʻunaue ʻana i (5 + 6i) me (2 + i) he 3.2 + 1.4i.
Kūkākūkā Hou: Modulus a me Conjugate o nā Helu Paʻakikī
Laʻana Nīnau 5
E huli i ka modulus a me ka conjugate o ka helu paʻakikī \(z = 3 + 4i\).
Kūkākūkā:
ʻO ke modulus o ka helu paʻakikī \(z = a + bi\):
\[
|z| = \sqrt{a^2 + b^2}
\]
No \(z = 3 + 4i\):
\[
|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
ʻO ka hui pū ʻana o ka helu paʻakikī \(z = a + bi\) ʻo \(z^ = a – bi\).
No \(z = 3 + 4i\):
\[
z^ = 3 – 4i
\]
No laila, ʻo ka modulus o \(3 + 4i\) he 5, a ʻo kona hui pū ʻana he \(3 – 4i\).
Ka hopena
He kuleana koʻikoʻi ko nā helu paʻakikī i nā ʻano hana makemakika like ʻole a me nā noi loea. ʻO ka hoʻomaopopo ʻana i nā hana kumu ma nā helu paʻakikī, e like me ka hoʻohui, ka hoʻemi, ka hoʻonui, a me ka mahele, he mea nui ia no ka hoʻohana ʻana i kēia mau manaʻo i ka hoʻoponopono ʻana i nā pilikia paʻakikī. ʻO ka hoʻomaʻamaʻa ʻana i nā ʻano pilikia like ʻole, e like me nā mea i wehewehe ʻia ma luna, e kōkua i ka hoʻoikaika ʻana i kou ʻike a me kou mau mākau i ka hana ʻana me nā helu paʻakikī.
