Piv txwv ntawm Cov Lus Nug thiab Kev Sib Tham txog Kev Ua Haujlwm ntawm Cov Naj Npawb Nyuaj
Cov lej sib xyaw yog kev txuas ntxiv ntawm lub tswv yim ntawm cov lej tiag tiag kom suav nrog cov lej xav xwb. Daim ntawv dav dav ntawm cov lej sib xyaw yog a + bi, qhov twg a thiab b yog cov lej tiag tiag, thiab i yog ib chav xav xwb nrog lub cuab yeej i² = -1. Kev ua haujlwm ntawm cov lej sib xyaw suav nrog kev ntxiv, kev rho tawm, kev sib npaug, thiab kev faib. Tsab xov xwm no yuav muab ntau yam piv txwv teeb meem thiab kev sib tham rau ntau yam kev ua haujlwm ntawm cov lej sib xyaw.
Kev Ntxiv thiab Kev Rho Tawm Cov Naj Npawb Uas Muaj Ntau Yam
Piv txwv lus nug 1
Ntxiv cov lej sib xyaw ua ke hauv qab no: (3 + 4i) thiab (1 + 2i).
Kev Sib Tham:
Kev ntxiv cov lej nyuaj yog ua tiav los ntawm kev ntxiv lawv cov feem tiag tiag thiab cov feem xav sib cais.
\[ (3 + 4i) + (1 + 2i) = (3 + 1) + (4i + 2i) = 4 + 6i \]
Yog li, qhov tshwm sim ntawm kev ntxiv (3 + 4i) thiab (1 + 2i) yog 4 + 6i.
Piv txwv lus nug 2
Rho tus lej nyuaj (2 + 5i) ntawm (6 + 3i).
Kev Sib Tham:
Kev rho tawm cov lej nyuaj yog ua los ntawm kev rho tawm qhov tiag tiag thiab qhov xav sib cais.
\[ (6 + 3i) – (2 + 5i) = (6 – 2) + (3i – 5i) = 4 – 2i \]
Yog li ntawd, qhov tshwm sim ntawm kev rho tawm (2 + 5i) ntawm (6 + 3i) yog 4 - 2i.
Kev Sib Npaug ntawm Cov Naj Npawb Sib Xyaws
Piv txwv lus nug 3
Muab cov lej sib xyaw ua ke no sib npaug: (2 + 3i) thiab (4 + i).
Kev Sib Tham:
Kev sib npaug ntawm cov lej nyuaj yog ua tiav los ntawm kev siv kev faib tawm lossis kev npaj ua raws cai, zoo ib yam li kev sib npaug ob lub binomials hauv algebra ib txwm muaj.
\[
(2 + 3i) \cdot (4 + i) = 2 \cdot 4 + 2 \cdot i + 3i \cdot 4 + 3i \cdot i
\]
Tom qab ntawd peb xam kom meej:
\[
= 8 + 2i + 12i + 3i^2
\]
Vim tias \( i^2 = -1 \):
\[
= 8 + 14i + 3(-1)
\]
\[
= 8 + 14i – 3
\]
\[
= 5+14i
\]
Yog li ntawd, qhov tshwm sim ntawm kev sib npaug (2 + 3i) thiab (4 + i) yog 5 + 14i.
Kev Faib Cov Naj Npawb Sib Xyaws
Piv txwv lus nug 4
Faib cov lej nyuaj no: (5 + 6i) los ntawm (2 + i).
Kev Sib Tham:
Kev faib cov lej nyuaj siv cov conjugate ntawm tus denominator. Cov conjugate ntawm \(2 + i\) yog \(2 - i\).
Peb muab tus lej thiab tus lej denominator sib npaug los ntawm cov conjugate ntawm cov denominators:
\[
\frac{5 + 6i}{2 + i} \cdot \frac{2 – i}{2 – i}
\]
Tam sim no peb xam tus lej thiab tus lej faib sib cais:
\[
= \frac{(5 + 6i) \cdot (2 – i)}{(2 + i) \cdot (2 – i)}
\]
Kev sib npaug ntawm cov denominators:
\[
(2 + i) \cdot (2 – i) = 2^2 – i^2 = 4 – (-1) = 4 + 1 = 5
\]
Kev sib npaug ntawm cov lej suav:
\[
(5 + 6i) \cdot (2 - i) = 5 \cdot 2 + 5 \cdot (-i) + 6i \cdot 2 + 6i \cdot (-i)
= 10 – 5i + 12i – 6i^2
= 10 + 7i – 6(-1)
= 10 + 7i + 6
= 16+7i
\]
Yog li, qhov kev faib tawm yog:
\[
= \frac{16+7i}{5} = \frac{16}{5} + \frac{7i}{5} = 3.2 + 1.4i
\]
Yog li ntawd, qhov tshwm sim ntawm kev faib (5 + 6i) los ntawm (2 + i) yog 3.2 + 1.4i.
Kev Sib Tham Ntxiv: Modulus thiab Conjugate ntawm Cov Naj Npawb Sib Xyaws
Piv txwv lus nug 5
Nrhiav tus modulus thiab conjugate ntawm tus lej complex \(z = 3 + 4i\).
Kev Sib Tham:
Tus modulus ntawm tus lej complex \(z = a + bi\) yog:
\[
|z| = \sqrt{a^2 + b^2}
\]
Rau \(z = 3 + 4i\):
\[
|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
Tus lej sib txuas ntawm tus lej nyuaj \(z = a + bi\) yog \(z^ = a – bi\).
Rau \(z = 3 + 4i\):
\[
z^ = 3 – 4i
\]
Yog li ntawd, tus modulus ntawm \(3 + 4i\) yog 5, thiab nws cov conjugate yog \(3 – 4i\).
Xaus
Cov lej nyuaj ua lub luag haujlwm tseem ceeb hauv ntau yam kev suav lej thiab kev siv tshuab. Kev nkag siab txog cov haujlwm yooj yim ntawm cov lej nyuaj, xws li kev ntxiv, kev rho tawm, kev sib npaug, thiab kev faib, yog qhov tseem ceeb rau kev siv cov tswv yim no hauv kev daws cov teeb meem nyuaj dua. Kev xyaum ntau hom teeb meem, xws li cov uas tau piav qhia saum toj no, yuav pab txhawb koj txoj kev nkag siab thiab kev txawj ntse hauv kev ua haujlwm nrog cov lej nyuaj.
