Nā nīnau hoʻohālike e kūkākūkā ana i nā helu paʻakikī
ʻO nā helu paʻakikī kahi kumuhana i ʻike pinepine ʻia ma ka makemakika ma nā pae kula kiʻekiʻe a me ke kulanui. ʻElua ʻāpana o nā helu paʻakikī: kahi ʻāpana maoli a me kahi ʻāpana hoʻokalakupua. Ma ka hoʻohana ʻana i ka hōʻailona maʻamau, ua kākau ʻia kahi helu paʻakikī ʻo \( z = a + bi \), kahi ʻo \( a \) a me \( b \) he mau helu maoli, a ʻo \( i \) ka ʻāpana hoʻokalakupua me ka waiwai \( i^2 = -1 \). E uhi kēia ʻatikala i kekahi mau laʻana a me kā lākou kūkākūkā e pili ana i nā helu paʻakikī, mai nā hana kumu a hiki i nā noi i ka hoʻoponopono pilikia.
Nā Nīnau Laʻana a me ke Kūkākūkā
1. Hoʻohui a me ka Hoʻemi ʻana o nā Helu Paʻakikī
Nīnau 1
E waiho \( z_1 = 3 + 4i \) a me \( z_2 = 1 – 2i \). E helu \( z_1 + z_2 \) a me \( z_1 – z_2 \).
Pahana
No ka hoʻohui a unuhi paha i nā helu paʻakikī, hoʻohana wale mākou i ka ʻāpana maoli me ka mea maoli a me ka ʻāpana hoʻokalakupua me ka hoʻokalakupua.
Hoʻohui:
\[
z_1 + z_2 = (3 + 4i) + (1 – 2i) = (3 + 1) + (4i – 2i) = 4 + 2i
\]
Hoʻemi:
\[
z_1 – z_2 = (3 + 4i) – (1 – 2i) = (3 – 1) + (4i + 2i) = 2 + 6i
\]
No laila, \( z_1 + z_2 = 4 + 2i \) a me \( z_1 – z_2 = 2 + 6i \).
2. Hoʻonui ʻia o nā helu paʻakikī
Nīnau 2
E helu i ka huahana o \( z_1 = 2 + 3i \) e \( z_2 = 4 – i \).
Pahana
No ka hoʻonui ʻana i ʻelua mau helu paʻakikī, hoʻohana mākou i ka waiwai hoʻolaha o ka algebra:
\[
z_1 \cdot z_2 = (2 + 3i)(4 – i)
\]
Hoʻonui mākou i kēlā me kēia ʻāpana:
\[
2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i)
\]
\[
= 8 – 2i + 12i – 3i^2
\]
No ka mea, \( i^2 = -1 \), a laila:
\[
= 8 – 2i + 12i + 3 = 11 + 10i
\]
No laila, ʻo ka huahana \( z_1 \cdot z_2 \) ʻo \( 11 + 10i \).
3. Ka Māhele ʻana o nā Helu Paʻakikī
Nīnau 3
E helu i ka quotient o \( z_1 = 3 + 4i \) e \( z_2 = 1 – i \).
Pahana
No ka puʻunaue ʻana i kahi helu paʻakikī, hoʻonui mākou i ka numerator a me ka denominator e ka conjugate o ka denominator o ka helu paʻakikī. ʻO ka conjugate o \( 1 – i \) ʻo \( 1 + i \).
\[
\frac{3 + 4i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(3 + 4i)(1 + i)}{(1 – i)(1 + i)}
\]
E helu mua kākou i ka denominator:
\[
(1 – i)(1 + i) = 1 – i^2 = 1 – (-1) = 2
\]
I kēia manawa ke helu nei mākou i ka numerator:
\[
(3 + 4i)(1 + i) = 3 + 3i + 4i + 4i^2 = 3 + 7i + 4(-1) = 3 + 7i – 4 = -1 + 7i
\]
No laila, ʻo ka hopena:
\[
\frac{-1 + 7i}{2} = -\frac{1}{2} + \frac{7}{2}i
\]
4. Modulus a me ka hoʻopaʻapaʻa o nā helu paʻakikī
Nīnau 4
E hoʻoholo i ka modulus a me ka hoʻopaʻapaʻa o \( z = 1 + i \).
Pahana
ʻO ke modulus o ka helu paʻakikī \( z = a + bi \) penei:
\[
|z| = \sqrt{a^2 + b^2}
\]
No \( z = 1 + i \), loaʻa iā mākou \( a = 1 \) a me \( b = 1 \):
\[
|z| = \sqrt{1^2 + 1^2} = \sqrt{2}
\]
ʻO ka hoʻopaʻapaʻa o kahi helu paʻakikī ʻo ia ke kihi \( \theta \) i hoʻokumu ʻia me ke axis maoli maikaʻi, i ana ʻia mai ke kumu a i ke kiko \( (a, b) \).
\[
\theta = \tan^{-1}\left(\frac{b}{a}\ʻākau)
\]
\[
\theta = \tan^{-1}(1) = \frac{\pi}{4}
\]
No laila, ʻo ka modulus o \( z = 1 + i \) ʻo \( \sqrt{2} \) a ʻo ka hoʻopaʻapaʻa ʻo \( \frac{\pi}{4} \).
5. ʻAno Exponential a me ke ʻano Euler
Nīnau 5
E hoʻololi i ka helu paʻakikī \( z = 1 + i \) i ke ʻano exponential.
Pahana
ʻAno exponential o nā helu paʻakikī e hoʻohana ana i ke ʻano hana a Euler:
\[
z = re^{i\theta}
\]
ʻO kahi ʻo \( r \) ka modulus a ʻo \( \theta \) ka hoʻopaʻapaʻa. Mai ke kūkākūkā mua, ʻike mākou:
\[
r = \sqrt{2}, \quad \theta = \frac{\pi}{4}
\]
No laila, ʻo ke ʻano exponential penei:
\[
z = \sqrt{2}e^{i\pi/4}
\]
6. Nā Aʻa o nā Helu Paʻakikī
Nīnau 6
E huli i nā aʻa huinahā o ka helu paʻakikī \( z = -1 \).
Pahana
Hiki ke loaʻa nā aʻa huinahā o nā helu paʻakikī me ka hoʻohana ʻana i ke ʻano polar a i ʻole exponential. Hoʻololi mākou iā \( z = -1 \) i ke ʻano exponential:
\[
z = -1 = e^{i\pi}
\]
Hiki ke kākau ʻia ke aʻa huinahā o \( e^{i\pi} \) penei:
\[
z_k = \sqrt{r} \cdot e^{i(\theta + 2k\pi)/n}
\]
Me \( r = 1 \), \( \theta = \pi \), \( n = 2 \), a me \( k = 0, 1 \):
\[
z_0 = e^{i(\pi + 2 \cdot 0 \cdot \pi)/2} = e^{i\pi/2} = i
\]
\[
z_1 = e^{i(\pi + 2 \cdot 1 \cdot \pi)/2} = e^{i3\pi/2} = -i
\]
No laila, ʻo nā aʻa huinahā o \( -1 \) ʻo \( i \) a me \( -i \).
7. Nā Hoʻohana ma nā Hoʻohālikelike Quadratic
Nīnau 7
E hoʻoponopono i ka hoohalike quadratic \( z^2 + 4z + 13 = 0 \).
Pahana
Hiki iā mākou ke hoʻohana i ke ʻano quadratic:
\[
z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
No ke kaulike \( z^2 + 4z + 13 = 0 \):
\[
a = 1, b = 4, c = 13
\]
\[
z = \frac{-4 \pm \sqrt{16 – 52}}{2 \cdot 1} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i
\]
No laila, ʻo nā hopena o \( z^2 + 4z + 13 = 0 \) ʻo \( z = -2 + 3i \) a me \( z = -2 – 3i \).
Ka hopena
He manaʻo makemakika ākea loa nā helu paʻakikī me nā noi he nui. Ma ka hoʻomaopopo ʻana i nā hana kumu e like me ka hoʻohui, ka hoʻemi, ka hoʻonui, a me ka mahele ʻana, a me ke ʻano o ka helu ʻana i ka modulus a me ka hoʻopaʻapaʻa, hiki iā mākou ke hoʻoponopono i nā pilikia like ʻole e pili ana i nā helu paʻakikī. Manaʻolana, e kōkua nā laʻana ma luna iā ʻoe e hoʻomaopopo maikaʻi a hoʻopaʻa i kēia kumuhana.
