Su'aalo Tusaale ah oo Ka Hadlaya Tirooyinka Adag
Tirooyinka isku dhafan waa mowduuc si joogto ah looga hadlo xisaabta heerarka dugsiga sare iyo kulliyadda labadaba. Tirooyinka isku dhafan waxay ka kooban yihiin laba qaybood: qayb dhab ah iyo qayb mala-awaal ah. Iyadoo la adeegsanayo calaamadaynta caadiga ah, tiro isku dhafan waxaa loo qoraa sida \( z = a + bi \), halkaas oo \( a \) iyo \( b \) ay yihiin tirooyin dhab ah, iyo \( i \) waa cutubka mala-awaalka ah ee leh hantida \( i^2 = -1 \). Maqaalkani wuxuu dabooli doonaa tusaalooyin dhowr ah iyo dooddooda ku saabsan tirooyinka isku dhafan, laga bilaabo hawlgallada aasaasiga ah ilaa codsiyada xallinta mashaakilaadka.
Su'aalo iyo Doodo Tusaale ah
1. Isugeynta iyo Kala-goynta Tirooyinka Adag
Su'aal 1aad
U daa \ ( z_1 = 3 + 4i \) iyo \ ( z_2 = 1 - 2i \). Xisaabi \( z_1 + z_2 \) iyo \( z_1 - z_2 \).
Dood
Si aan isugu darno ama u kala jarno tirooyinka adag, waxaan si fudud u isticmaalnaa qaybta dhabta ah iyadoo la adeegsanayo dhabta ah iyo qaybta male-awaalka ah iyadoo la adeegsanayo male-awaalka.
Ku darid:
\[
z_1 + z_2 = (3 + 4i) + (1 – 2i) = (3 + 1) + (4i – 2i) = 4 + 2i
\]
Kala-goynta:
\[
z_1 – z_2 = (3 + 4i) – (1 – 2i) = (3 – 1) + (4i + 2i) = 2 + 6i
\]
Haddaba, \ ( z_1 + z_2 = 4 + 2i \) iyo \( z_1 – z_2 = 2 + 6i \).
2. Isku-dhufashada Tirooyinka Isku-dhafan
Su'aal 2aad
Xisaabi natiijada \( z_1 = 2 + 3i \) adoo isticmaalaya \( z_2 = 4 – i \).
Dood
Si aan u dhufanno laba tiro oo isku dhafan, waxaan isticmaalnaa sifada qaybinta ee aljabrada:
\[
z_1 \cdot z_2 = (2 + 3i)(4 – i)
\]
Waxaan ku dhufaneynaa qayb kasta:
\[
2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i)
\]
\[
= 8 – 2i + 12i – 3i^2
\]
Tan iyo \( i^2 = -1 \), markaa:
\[
= 8 – 2i + 12i + 3 = 11 + 10i
\]
Markaa, badeecaddu \( z_1 \cdot z_2 \) waa \( 11 + 10i \).
3. Qaybinta Tirooyinka Isku-dhafan
Su'aal 3aad
Xisaabi wadarta \( z_1 = 3 + 4i \) adoo isticmaalaya \( z_2 = 1 – i \).
Dood
Si aan u qaybinno tiro isku dhafan, waxaan ku dhufanaynaa tirada iyo hooseeyaha isku-xidhka hooseeyaha ee tirada isku-dhafan. Isku-xidhka \( 1 – i \) waa \( 1 + i \).
\[
\frac{3 + 4i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(3 + 4i)(1 + i)}{(1 – i)(1 + i)}
\]
Aan marka hore xisaabino hooseeyaha:
\[
(1 – i)(1 + i) = 1 – i^2 = 1 – (-1) = 2
\]
Hadda waxaan xisaabinaynaa tiro-ururiyaha:
\[
(3 + 4i) (1 + i) = 3 + 3i + 4i + 4i^2 = 3 + 7i + 4 (-1) = 3 + 7i – 4 = -1 + 7i
\]
Markaas, natiijadu waa:
\[
\frac{-1 + 7i}{2} = -\frac{1}{2} + \frac{7}{2}i
\]
4. Modulus iyo Doodda Tirooyinka Adag
Su'aal 4aad
Go'aami module-ka iyo doodda \( z = 1 + i \).
Dood
Modulus-ka tirada isku dhafan \( z = a + bi \) waa:
\[
|z| = \sqrt{a^2 + b^2}
\]
Wixii \( z = 1 + i \), waxaan haysannaa \( a = 1 \) iyo \( b = 1 \):
\[
|z| = \sqrt{1^2 + 1^2} = \sqrt{2}
\]
Doodda tiro isku dhafan waa xagasha \( \theta \) oo lagu sameeyay dhidibka dhabta ah ee togan, oo laga cabbiray asalka ilaa barta \( (a, b) \).
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
\[
\theta = \tan^{-1}(1) = \frac{\pi}{4}
\]
Markaa, module-ka \( z = 1 + i \) waa \( \sqrt{2} \) doodduna waa \( \frac{\pi}{4} \).
5. Qaabka Jilitaanka iyo Qaabka Euler
Su'aal 5aad
U beddel tirada isku dhafan \( z = 1 + i \) qaab jibbaaran.
Dood
Qaabka jibbaarada ee tirooyinka isku dhafan iyadoo la adeegsanayo qaacidada Euler:
\[
z = re^{i\theta}
\]
Halka \( r \) uu yahay modules-ka iyo \( \theta \) ay tahay dooddu. Dooddii hore, waxaan ka ognahay in:
\[
r = \sqrt{2}, \quad \theta = \frac{\pi}{4}
\]
Haddaba, qaabka jibbaaran waa:
\[
z = \sqrt{2}e^{i\pi/4}
\]
6. Xididdada Tirooyinka Isku-dhafan
Su'aal 6aad
Soo hel xididdada laba jibbaaran ee tirada isku dhafan \( z = -1 \).
Dood
Xididdada labajibbaaran ee tirooyinka isku dhafan waxaa laga heli karaa iyadoo la adeegsanayo qaabka polar ama exponential. Waxaan u beddelnaa \( z = -1 \) qaab exponential ah:
\[
z = -1 = e^{i\pi}
\]
Xididka labajibbaaran ee \( e^{i\pi} \) waxaa loo qori karaa sidan:
\[
z_k = \sqrt{r} \cdot e^{i(\theta + 2k\pi)/n}
\]
Iyadoo \( r = 1 \), \( \theta = \pi \), \( n = 2 \), iyo \( k = 0, 1 \):
\[
z_0 = e^{i(\pi +2 \cdot 0 \cdot \pi)/2} = e^{i\pi/2} = i
\]
\[
z_1 = e^{i(\pi +2 \cdot 1 \cdot \pi)/2} = e^{i3\pi/2} = -i
\]
Haddaba, xididdada laba jibbaaran ee \( -1 \) waa \( i \) iyo \( -i \).
7. Codsiyada Isle'egyada Labajibbaaran
Su'aal 7aad
Xalli isle'egta labajibbaaran \( z^2 + 4z + 13 = 0 \).
Dood
Waxaan isticmaali karnaa qaacidada labajibbaaran:
\[
z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
Isleegta \( z^2 + 4z + 13 = 0 \):
\[
a = 1, b = 4, c = 13
\]
\[
z = \frac{-4 \pm \sqrt{16 – 52}}{2 \cdot 1} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i
\]
Markaa, xalalka \( z^2 + 4z + 13 = 0 \) waa \( z = -2 + 3i \) iyo \( z = -2 – 3i \).
Gabagabo
Tirooyinka isku dhafan waa fikrad xisaabeed oo aad u ballaaran oo leh codsiyo badan. Annagoo fahanayna hawlgallada aasaasiga ah sida isku-darka, kala-goynta, isku-dhufashada, iyo qaybinta, iyo sidoo kale sida loo xisaabiyo modules-ka iyo doodda, waxaan xallin karnaa dhibaatooyin kala duwan oo ku lug leh tirooyinka isku dhafan. Waxaan rajeyneynaa, tusaalooyinka kor ku xusan inay kaa caawin doonaan inaad si fiican u fahamto oo aad u barato mowduucan.
