Mehlala ea Lipotso Tse Buisanang ka Linomoro Tse Rarahaneng
Linomoro tse rarahaneng ke sehlooho se atisang ho tluwa ho sona dipalong tsa dipalo maemong a sekolo se phahameng le a koleche. Dinomoro tse rarahaneng di na le dikarolo tse pedi: karolo ya sebele le karolo e inahaneloang. Ho sebediswa mongolo o tlwaelehileng, nomoro e rarahaneng e ngolwa e le \( z = a + bi \), moo \( a \) le \( b \) e leng dinomoro tsa nnete, mme \( i \) ke yuniti e inahaneloang e nang le thepa \( i^2 = -1 \). Sengoloa sena se tla akaretsa mehlala e mmalwa le puisano ya yona mabapi le dinomoro tse rarahaneng, ho tloha tshebetsong ya motheo ho ya ditshebedisong tsa ho rarolla mathata.
Lipotso tsa Mehlala le Puisano
1. Ho eketsa le ho tlosa linomoro tse rarahaneng
Potso ea 1
Tlohela \( z_1 = 3 + 4i \) le \( z_2 = 1 – 2i \). Bala \( z_1 + z_2 \) le \( z_1 – z_2 \).
Puisano
Ho eketsa kapa ho tlosa dinomoro tse rarahaneng, re mpa re sebedisa karolo ya nnete ka ya sebele le karolo e inahanelwang ka ya boiqapelo.
Ho eketsa:
\[
z_1 + z_2 = (3 + 4i) + (1 – 2i) = (3 + 1) + (4i – 2i) = 4 + 2i
\]
Ho tlosa:
\[
z_1 – z_2 = (3 + 4i) – (1 – 2i) = (3 – 1) + (4i + 2i) = 2 + 6i
\]
Kahoo, \( z_1 + z_2 = 4 + 2i \) le \( z_1 – z_2 = 2 + 6i \).
2. Katiso ea Linomoro tse Rarahaneng
Potso ea 2
Bala sehlahisoa sa \( z_1 = 2 + 3i \) ka \( z_2 = 4 – i \).
Puisano
Ho atisa dinomoro tse pedi tse rarahaneng, re sebedisa thepa ya kabo ya algebra:
\[
z_1 \cdot z_2 = (2 + 3i)(4 – i)
\]
Re atisa karolo ka 'ngoe:
\[
2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i)
\]
\[
= 8 – 2i + 12i – 3i^2
\]
Kaha \( i^2 = -1 \), joale:
\[
= 8 – 2i + 12i + 3 = 11 + 10i
\]
Kahoo, sehlahisoa \( z_1 \cdot z_2 \) ke \( 11 + 10i \).
3. Karohano ea Linomoro tse Rarahaneng
Potso ea 3
Bala palo ea \( z_1 = 3 + 4i \) ka \( z_2 = 1 – i \).
Puisano
Ho arola nomoro e rarahaneng, re atisa nomoro le denominator ka conjugate ea denominator ea nomoro e rarahaneng. Conjugate ea \( 1 - i \) ke \( 1 + i \).
\[
\frac{3 + 4i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(3 + 4i)(1 + i)}{(1 – i)(1 + i)}
\]
A re baleng denominator pele:
\[
(1 – i)(1 + i) = 1 – i^2 = 1 – (-1) = 2
\]
Jwale re bala nomoro:
\[
(3 + 4i) (1 + i) = 3 + 3i + 4i + 4i^2 = 3 + 7i + 4(-1) = 3 + 7i – 4 = -1 + 7i
\]
Kahoo, sephetho ke:
\[
\frac{-1 + 7i}{2} = -\frac{1}{2} + \frac{7}{2}i
\]
4. Modulus le Khang ea Linomoro tse Rarahaneng
Potso ea 4
Fumana modulus le ngangisano ea \( z = 1 + i \).
Puisano
Modulus ea nomoro e rarahaneng \( z = a + bi \) ke:
\[
|z| = \sqrt{a^2 + b^2}
\]
Bakeng sa \( z = 1 + i \), re na le \( a = 1 \) le \( b = 1 \):
\[
|z| = \sqrt{1^2 + 1^2} = \sqrt{2}
\]
Kgang ya nomoro e rarahaneng ke sekhutlo \( \theta \) se bopilweng ka mothapo wa nnete o motle, o lekantsweng ho tloha tshimolohong ho ya ntlheng \( (a, b) \).
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
\[
\theta = \tan^{-1}(1) = \frac{\pi}{4}
\]
Kahoo, modulus ea \( z = 1 + i \) ke \( \sqrt{2} \) 'me khang ke \( \frac{\pi}{4} \).
5. Sebopeho sa Exponential le Paterone ea Euler
Potso ea 5
Fetola nomoro e rarahaneng \( z = 1 + i \) hore e be sebopeho sa exponential.
Puisano
Sebopeho sa tlhaloso sa linomoro tse rarahaneng se sebelisang foromo ea Euler:
\[
z = re^{i\theta}
\]
Moo \( r \) e leng modulus le \( \theta \) e leng ngangisano. Ho tsoa puisanong e fetileng, rea tseba hore:
\[
r = \sqrt{2}, \quad \theta = \frac{\pi}{4}
\]
Kahoo, foromo ea exponential ke:
\[
z = \sqrt{2}e^{i\pi/4}
\]
6. Metso ea Lipalo tse Rarahaneng
Potso ea 6
Fumana metso e sekwere ya nomoro e rarahaneng \( z = -1 \).
Puisano
Metso e sekwere ya dinomoro tse rarahaneng e ka fumanwa ka ho sebedisa sebopeho sa polar kapa sa exponential. Re fetolela \( z = -1 \) hore e be sebopeho sa exponential:
\[
z = -1 = e^{i\pi}
\]
Motso o sekwere wa \( e^{i\pi} \) o ka ngolwa tjena:
\[
z_k = \sqrt{r} \cdot e^{i(\theta + 2k\pi)/n}
\]
Ka \( r = 1 \), \( \theta = \pi \), \( n = 2 \), le \( k = 0, 1 \):
\[
z_0 = e^{i(\pi + 2 \cdot 0 \cdot \pi)/2} = e^{i\pi/2} = i
\]
\[
z_1 = e^{i(\pi + 2 \cdot 1 \cdot \pi)/2} = e^{i3\pi/2} = -i
\]
Kahoo, metso e sekwere ya \( -1 \) ke \( i \) le \( -i \).
7. Litšebeliso ho Quadratic Equations
Potso ea 7
Rarolla equation ea quadratic \( z^2 + 4z + 13 = 0 \).
Puisano
Re ka sebelisa foromo ea quadratic:
\[
z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
Bakeng sa equation \( z^2 + 4z + 13 = 0 \):
\[
a = 1, b = 4, c = 13
\]
\[
z = \frac{-4 \pm \sqrt{16 – 52}}{2 \cdot 1} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i
\]
Kahoo, ditharollo tsa \( z^2 + 4z + 13 = 0 \) ke \( z = -2 + 3i \) le \( z = -2 – 3i \).
Qetello
Linomoro tse rarahaneng ke khopolo-taba e pharaletseng haholo ea lipalo e nang le lits'ebetso tse ngata. Ka ho utloisisa ts'ebetso ea motheo joalo ka ho eketsa, ho tlosa, ho atisa le ho arola, hammoho le mokhoa oa ho bala modulus le ngangisano, re ka rarolla mathata a fapaneng a amanang le linomoro tse rarahaneng. Re tšepa hore mehlala e kaholimo e tla u thusa ho utloisisa le ho tseba sehlooho sena hamolemo.
