Tambayoyi Misali Game da Lambobi Masu Rikitarwa
Lambobi masu rikitarwa batu ne da ake yawan fuskanta a lissafi a matakan sakandare da na kwaleji. Lambobi masu rikitarwa sun ƙunshi sassa biyu: ɓangare na gaske da ɓangaren tunani. Ta amfani da alamun rubutu na al'ada, ana rubuta lamba mai rikitarwa kamar haka \( z = a + bi \), inda \( a \) da \( b \) lambobi ne na gaske, kuma \( i \) shine raka'ar tunani tare da mallakar \( i^2 = -1 \). Wannan labarin zai ƙunshi misalai da yawa da tattaunawarsu game da lambobi masu rikitarwa, daga ayyukan asali zuwa aikace-aikace a cikin warware matsaloli.
Tambayoyi da Tattaunawa Samfura
1. Ƙarawa da Ragewa na Lambobi Masu Hadaka
Tambaya ta 1
Bari \ ( z_1 = 3 + 4i \) da (z_2 = 1 - 2i \). Yi lissafin \(z_1 + z_2 \) da \(z_1 - z_2 \).
Tattaunawa
Domin ƙara ko cire lambobi masu rikitarwa, kawai muna amfani da ainihin ɓangaren da ainihin ɓangaren da kuma ɓangaren da aka ƙirƙira da tunanin.
Ƙari:
\[
z_1 + z_2 = (3 + 4i) + (1 – 2i) = (3 + 1) + (4i – 2i) = 4 + 2i
\]
Ragewa:
\[
z_1 – z_2 = (3 + 4i) – (1 – 2i) = (3 – 1) + (4i + 2i) = 2 + 6i
\]
Don haka, (z_1 + z_2 = 4 + 2i \) da \ (z_1 – z_2 = 2 + 6i \).
2. Yawan Lambobi Masu Hadaka
Tambaya ta 2
Lissafi samfurin \( z_1 = 2 + 3i \) ta hanyar \( z_2 = 4 – i \).
Tattaunawa
Don ninka lambobi biyu masu rikitarwa, muna amfani da sifar rarrabawa ta algebra:
\[
z_1 \cdot z_2 = (2 + 3i)(4 – i)
\]
Muna ninka kowanne bangare:
\[
2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i)
\]
\[
= 8 – 2i + 12i – 3i^2
\]
Tunda \(i^2 = -1 \), to:
\[
= 8 – 2i + 12i + 3 = 11 + 10i
\]
Don haka, samfurin \( z_1 \cdot z_2 \) shine \( 11 + 10i \).
3. Raba Lambobi Masu Hadaka
Tambaya ta 3
Lissafa jimlar \( z_1 = 3 + 4i \) ta hanyar \( z_2 = 1 – i \).
Tattaunawa
Domin raba lamba mai rikitarwa, muna ninka mai ƙidaya da mai ƙidaya ta hanyar haɗa mai ƙidayar lambar mai rikitarwa. Haɗin \( 1 – i \) shine \( 1 + i \).
\[
\frac{3 + 4i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(3 + 4i)(1 + i)}{(1 – i)(1 + i)}
\]
Bari mu fara lissafin ma'aunin rabo:
\[
(1 – i)(1 + i) = 1 – i^2 = 1 – (-1) = 2
\]
Yanzu mun ƙididdige mai ƙidaya:
\[
(3 + 4i) (1 + i) = 3 + 3i + 4i + 4i^2 = 3 + 7i + 4 (-1) = 3 + 7i – 4 = -1 + 7i
\]
Don haka, sakamakon shine:
\[
\frac{-1 + 7i}{2} = -\frac{1}{2} + \frac{7}{2}i
\]
4. Modulus da Hujjar Lambobi Masu Hadaka
Tambaya ta 4
Kayyade tsarin da hujjar \( z = 1 + i \).
Tattaunawa
Modulus na lambar hadaddun \( z = a + bi \) shine:
\[
|z| = \sqrt{a^2 + b^2}
\]
Ga \( z = 1 + i \), muna da \( a = 1 \) da \( b = 1 \):
\[
|z| = \sqrt{1^2 + 1^2} = \sqrt{2}
\]
Hujjar lamba mai rikitarwa ita ce kusurwar \( \theta \) da aka samar tare da madaidaicin axis na gaske, wanda aka auna daga asali zuwa wurin \( (a, b) \).
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
\[
\theta = \tan^{-1}(1) = \frac{\pi}{4}
\]
Don haka, tsarin \( z = 1 + i \) shine \( \sqrt{2} \) kuma hujjar ita ce \( \frac{\pi}{4} \).
5. Siffar Expo da Tsarin Euler
Tambaya ta 5
Maida lambar hadaddun \( z = 1 + i \) zuwa siffar mai faɗi.
Tattaunawa
Siffar lambobi masu rikitarwa ta amfani da dabarar Euler:
\[
z = re^{i\theta}
\]
Inda \( r \) shine modulus da \( \theta \) shine hujja. Daga tattaunawar da ta gabata, mun san cewa:
\[
r = \sqrt{2}, \quad \theta = \frac{\pi}{4}
\]
Don haka, siffar exponential ita ce:
\[
z = \sqrt{2}e^{i\pi/4}
\]
6. Tushen Lambobi Masu Hadaka
Tambaya ta 6
Nemo tushen murabba'in lambar hadaddun \( z = -1 \).
Tattaunawa
Ana iya samun tushen murabba'in lambobi masu rikitarwa ta amfani da siffar polar ko exponential. Muna mayar da \( z = -1 \) zuwa siffar exponential:
\[
z = -1 = e^{i\pi}
\]
Tushen murabba'in \(e^{i\pi} \) za a iya rubuta shi kamar haka:
\[
z_k = \sqrt{r} \cdot e^{i(\theta + 2k\pi)/n}
\]
Tare da \( r = 1 \), \( \theta = \pi \), \( n = 2 \), da \( k = 0, 1 \):
\[
z_0 = e^{i(\pi + 2 \cdot 0 \cdot \pi)/2} = e^{i\pi/2} = i
\]
\[
z_1 = e^{i(\pi + 2 \cdot 1 \cdot \pi)/2} = e^{i3\pi/2} = -i
\]
Don haka, tushen murabba'in \( -1 \) sune \( i \) da \( -i \).
7. Aikace-aikace a cikin Lissafin Huɗu
Tambaya ta 7
Warware lissafin kwata-kwata \( z^2 + 4z + 13 = 0 \).
Tattaunawa
Za mu iya amfani da dabarar quadratic:
\[
z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
Ga lissafin \( z^2 + 4z + 13 = 0 \):
\[
a = 1, b = 4, c = 13
\]
\[
z = \frac{-4 \pm \sqrt{16 – 52}}{2 \cdot 1} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i
\]
Don haka, mafita na \( z^2 + 4z + 13 = 0 \) sune \( z = -2 + 3i \) da \( z = -2 – 3i \).
Kammalawa
Lambobi masu rikitarwa wata faffadan ra'ayi ne na lissafi tare da aikace-aikace da yawa. Ta hanyar fahimtar ayyukan asali kamar ƙari, ragi, ninkawa, da rabawa, da kuma yadda ake ƙididdige modulus da hujja, za mu iya magance matsaloli daban-daban da suka shafi lambobi masu rikitarwa. Da fatan, misalan da ke sama za su taimaka muku fahimtar da kuma fahimtar wannan batu sosai.
