Imibuzo Eyisibonelo Exoxa Ngezinombolo Eziyinkimbinkimbi
Izinombolo eziyinkimbinkimbi ziyisihloko esivame ukuhlangatshezwana naso ezibalweni emazingeni esikole samabanga aphezulu kanye nasekolishi. Izinombolo eziyinkimbinkimbi zinezingxenye ezimbili: ingxenye yangempela kanye nengxenye engokomfanekiso. Kusetshenziswa ukubhalwa okuvamile, inombolo eyinkimbinkimbi ibhalwa njengo-\( z = a + bi \), lapho \( a \) kanye no-\( b \) kuyizinombolo zangempela, kanti \( i \) iyunithi engokomfanekiso enesakhiwo \( i^2 = -1 \). Lesi sihloko sizomboza izibonelo eziningana kanye nengxoxo yazo mayelana nezinombolo eziyinkimbinkimbi, kusukela ekusebenzeni okuyisisekelo kuya ekusetshenzisweni ekuxazululeni izinkinga.
Imibuzo Eyisibonelo Nengxoxo
1. Ukuhlanganisa Nokususa Izinombolo Eziyinkimbinkimbi
umbuzo 1
Vumela \( z_1 = 3 + 4i \) kanye \( z_2 = 1 – 2i \). Bala \( z_1 + z_2 \) kanye \( z_1 – z_2 \).
Ingxoxo
Ukuze singeze noma sisuse izinombolo eziyinkimbinkimbi, simane sisebenzise ingxenye yangempela neyangempela kanye nengxenye ecatshangelwayo neyangempela.
Ukwengeza:
\[
z_1 + z_2 = (3 + 4i) + (1 – 2i) = (3 + 1) + (4i – 2i) = 4 + 2i
\]
Ukususa:
\[
z_1 – z_2 = (3 + 4i) – (1 – 2i) = (3 – 1) + (4i + 2i) = 2 + 6i
\]
Ngakho, \( z_1 + z_2 = 4 + 2i \) kanye \( z_1 – z_2 = 2 + 6i \).
2. Ukuphindaphinda Kwezinombolo Eziyinkimbinkimbi
umbuzo 2
Bala umkhiqizo we-\( z_1 = 2 + 3i \) ngo-\( z_2 = 4 – i \).
Ingxoxo
Ukuze siphindaphinde izinombolo ezimbili eziyinkimbinkimbi, sisebenzisa i-algebra enikeza ukwahlukanisa:
\[
z_1 \cdot z_2 = (2 + 3i)(4 – i)
\]
Siphindaphinda ingxenye ngayinye:
\[
2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i)
\]
\[
= 8 – 2i + 12i – 3i^2
\]
Kusukela ku-\( i^2 = -1 \), khona-ke:
\[
= 8 – 2i + 12i + 3 = 11 + 10i
\]
Ngakho-ke, umkhiqizo \( z_1 \cdot z_2 \) ngu \( 11 + 10i \).
3. Ukuhlukaniswa Kwezinombolo Eziyinkimbinkimbi
umbuzo 3
Bala isilinganiso sika-\( z_1 = 3 + 4i \) ngo-\( z_2 = 1 – i \).
Ingxoxo
Ukuze sihlukanise inombolo eyinkimbinkimbi, siphinda i-numerator kanye ne-denominator nge-conjugate ye-denominator yenombolo eyinkimbinkimbi. I-conjugate ye-\( 1 – i \) ingu-\( 1 + i \).
\[
\frac{3 + 4i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(3 + 4i)(1 + i)}{(1 – i)(1 + i)}
\]
Ake sibale i-denominator kuqala:
\[
(1 – i)(1 + i) = 1 – i^2 = 1 – (-1) = 2
\]
Manje sibala inombolo:
\[
(3 + 4i)(1 + i) = 3 + 3i + 4i + 4i^2 = 3 + 7i + 4(-1) = 3 + 7i – 4 = -1 + 7i
\]
Ngakho-ke, umphumela uba:
\[
\frac{-1 + 7i}{2} = -\frac{1}{2} + \frac{7}{2}i
\]
4. I-Modulus kanye ne-Agument yezinombolo eziyinkimbinkimbi
umbuzo 4
Nquma i-modulus kanye ne-agumenti ka-\( z = 1 + i \).
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I-modulus yenombolo eyinkimbinkimbi \( z = a + bi \) ithi:
\[
|z| = \sqrt{a^2 + b^2}
\]
Ku-\( z = 1 + i \), sine-\( a = 1 \) kanye ne-\( b = 1 \):
\[
|z| = \sqrt{1^2 + 1^2} = \sqrt{2}
\]
Impikiswano yenombolo eyinkimbinkimbi yi-engeli \( \theta \) eyakhiwe nge-axis yangempela eqondile, elinganiswa kusukela ekuqaleni kuya ephuzwini \( (a, b) \).
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
\[
\theta = \tan^{-1}(1) = \frac{\pi}{4}
\]
Ngakho-ke, i-modulus ka-\( z = 1 + i \) ingu-\( \sqrt{2} \) kanti impikiswano ingu-\( \frac{\pi}{4} \).
5. Ifomu Lokuchaza kanye Nephethini Lase-Euler
umbuzo 5
Guqula inombolo eyinkimbinkimbi \( z = 1 + i \) ibe yifomu le-exponential.
Ingxoxo
Ifomu le-Exponential lezinombolo eziyinkimbinkimbi kusetshenziswa ifomula ka-Euler:
\[
z = re^{i\theta}
\]
Lapho \( r \) kuyi-modulus kanye \( \theta \) kuyimpikiswano. Kusukela engxoxweni edlule, siyazi ukuthi:
\[
r = \sqrt{2}, \quad \theta = \frac{\pi}{4}
\]
Ngakho-ke, ifomu le-exponential lithi:
\[
z = \sqrt{2}e^{i\pi/4}
\]
6. Izimpande Zezinombolo Eziyinkimbinkimbi
umbuzo 6
Thola izimpande zesikwele zenombolo eyinkimbinkimbi \( z = -1 \).
Ingxoxo
Izimpande zesikwele zezinombolo eziyinkimbinkimbi zingatholakala kusetshenziswa ifomu le-polar noma le-exponential. Siguqula i-\( z = -1 \) ibe ifomu le-exponential:
\[
z = -1 = e^{i\pi}
\]
Impande yesikwele ye-\( e^{i\pi} \) ingabhalwa kanje:
\[
z_k = \sqrt{r} \cdot e^{i(\theta + 2k\pi)/n}
\]
Nge-\( r = 1 \), \( \theta = \pi \), \( n = 2 \), kanye ne-\( k = 0, 1 \):
\[
z_0 = e^{i(\pi + 2 \cdot 0 \cdot \pi)/2} = e^{i\pi/2} = i
\]
\[
z_1 = e^{i(\pi + 2 \cdot 1 \cdot \pi)/2} = e^{i3\pi/2} = -i
\]
Ngakho-ke, izimpande zesikwele ze-\( -1 \) ziyi-\( i \) kanye ne-\( -i \).
7. Izicelo ku-Quadratic Equations
umbuzo 7
Xazulula i-quadratic equation \( z^2 + 4z + 13 = 0 \).
Ingxoxo
Singasebenzisa ifomula ye-quadratic:
\[
z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
Ukuze uthole i-equation \( z^2 + 4z + 13 = 0 \):
\[
a = 1, b = 4, c = 13
\]
\[
z = \frac{-4 \pm \sqrt{16 – 52}}{2 \cdot 1} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i
\]
Ngakho-ke, izixazululo ze-\( z^2 + 4z + 13 = 0 \) ziyi-\( z = -2 + 3i \) kanye ne-\( z = -2 – 3i \).
Isiphetho
Izinombolo eziyinkimbinkimbi ziwumqondo obanzi kakhulu wezibalo onezinhlelo zokusebenza eziningi. Ngokuqonda imisebenzi eyisisekelo efana nokuhlanganisa, ukususa, ukuphindaphinda, nokuhlukanisa, kanye nendlela yokubala i-modulus kanye ne-argument, singaxazulula izinkinga ezahlukahlukene ezihilela izinombolo eziyinkimbinkimbi. Ngethemba ukuthi izibonelo ezingenhla zizokusiza ukuthi uqonde kangcono futhi ube nekhono kulesi sihloko.
