Ngā Tauira o ngā Pātai me ngā Kōrero mō ngā Mahi Tau Matatini
Ko ngā tau matatini he toronga o te ariā o ngā tau tūturu hei whakauru i ngā tau pohewa. Ko te āhua whānui o tētahi tau matatini ko a + bi, ko a me b he tau tūturu, ā, ko i he wae pohewa me te āhuatanga i² = -1. Ko ngā mahi i runga i ngā tau matatini ko te tāpiri, te tango, te whakarea, me te wehewehe. Ka whakaratohia e tēnei tuhinga ētahi tauira rapanga me ngā kōrero mō ngā mahi rerekē i runga i ngā tau matatini.
Te Tāpiri me te Tango i ngā Tau Matatini
Tauira Pātai 1
Tāpirihia ngā tau matatini e whai ake nei: (3 + 4i) me (1 + 2i).
Kōrero:
Ka taea te tāpiri i ngā tau matatini mā te tāpiri motuhake i ō rātou wāhanga tūturu me ō rātou wāhanga pohewa.
\[ (3 + 4i) + (1 + 2i) = (3 + 1) + (4i + 2i) = 4 + 6i \]
Nō reira, ko te hua o te tāpiri i te (3 + 4i) me te (1 + 2i) ko te 4 + 6i.
Tauira Pātai 2
Tangohia te tau matatini (2 + 5i) mai i (6 + 3i).
Kōrero:
Ka taea te tango i ngā tau matatini mā te tango motuhake i te wāhanga tūturu me te wāhanga pohewa.
\[ (6 + 3i) – (2 + 5i) = (6 – 2) + (3i – 5i) = 4 – 2i \]
Nō reira, ko te hua o te tango i te (2 + 5i) mai i te (6 + 3i) ko te 4 – 2i.
Te Whakarea o ngā Tau Uaua
Tauira Pātai 3
Whakareatia ngā tau matatini e whai ake nei: (2 + 3i) me (4 + i).
Kōrero:
Ka mahia te whakarea o ngā tau matatini mā te whakamahi i ngā tohatoha, i ngā whakaritenga ōkawa rānei, he rite ki te whakarea i ngā binomial e rua i roto i te arapūrei noa.
\[
(2 + 3i) \cdot (4 + i) = 2 \cdot 4 + 2 \cdot i + 3i \cdot 4 + 3i \cdot i
\]
Kātahi ka tatauhia e mātou i roto i ngā taipitopito:
\[
= 8 + 2i + 12i + 3i^2
\]
Mai i te mea \( i^2 = -1 \):
\[
= 8 + 14i + 3(-1)
\]
\[
= 8 + 14i – 3
\]
\[
= 5 + 14i
\]
Nō reira, ko te hua o te whakarea i te (2 + 3i) me te (4 + i) ko te 5 + 14i.
Te Wehewehenga o ngā Tau Uaua
Tauira Pātai 4
Wehea te tau matatini e whai ake nei: (5 + 6i) ki (2 + i).
Kōrero:
Te wehewehe i ngā tau matatini mā te whakamahi i te hononga o te tauwehenga. Ko te hononga o \(2 + i\) ko \(2 – i\).
Ka whakareatia e tātou te taupū me te taupū ki te hononga o ngā taupū:
\[
\frac{5 + 6i}{2 + i} \cdot \frac{2 – i}{2 – i}
\]
Inaianei ka tatauhia e tātou te taupū me te tauwehe motuhake:
\[
= \frac{(5 + 6i) \cdot (2 – i)}{(2 + i) \cdot (2 – i)}
\]
Te whakarea o ngā taupū:
\[
(2 + i) \cdot (2 – i) = 2^2 – i^2 = 4 – (-1) = 4 + 1 = 5
\]
Te whakarea o ngā taupū:
\[
(5 + 6i) \cdot (2 – i) = 5 \cdot 2 + 5 \cdot (-i) + 6i \cdot 2 + 6i \cdot (-i)
= 10 – 5i + 12i – 6i^2
= 10 + 7i – 6(-1)
= 10 + 7i + 6
= 16 + 7i
\]
Nā, ko te wehenga ko:
\[
= \frac{16 + 7i}{5} = \frac{16}{5} + \frac{7i}{5} = 3.2 + 1.4i
\]
Nō reira, ko te hua o te wehewehe i te (5 + 6i) ki te (2 + i) ko te 3.2 + 1.4i.
Kōrero Tāpiri: Te Tauwehenga me te Hononga o ngā Tau Matatini
Tauira Pātai 5
Kimihia te modulus me te hononga o te tau matatini \(z = 3 + 4i\).
Kōrero:
Ko te modulus o te tau matatini \(z = a + bi\) ko:
\[
|z| = \sqrt{a^2 + b^2}
\]
Mō \(z = 3 + 4i\):
\[
|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
Ko te hononga o te tau matatini \(z = a + bi\) ko \(z^ = a – bi\).
Mō \(z = 3 + 4i\):
\[
z^ = 3 – 4i
\]
Nō reira, ko te modulus o \(3 + 4i\) he 5, ā, ko tōna hononga ko \(3 – 4i\).
Whakamutunga
He mea nui te mahi a ngā tau matatini i roto i te whānuitanga o ngā mara pāngarau me ngā tono hangarau. Ko te mārama ki ngā mahi taketake o ngā tau matatini, pērā i te tāpiri, te tango, te whakarea, me te wehewehe, he mea nui ki te whakamahi i ēnei ariā hei whakaoti rapanga uaua ake. Mā te whakaharatau i ngā momo rapanga rerekē, pērā i ngā mea kua whakaahuatia i runga ake nei, ka āwhina i te whakapakari i tō māramatanga me ō pūkenga ki te mahi me ngā tau matatini.
