Lanjutan Bilangan Kompleks: Operasi Aljabar Bilangan Kompleks (Matematika Tingkat Lanjut Kelas XI)

D. Operasi Aljabar Bilangan Kompleks

Perhatikan tabel berikut

$\begin{array}{|cll|}\hline \mathbf{\text{No}}& \mathbf{\text{Jenis Operasi}}& \mathbf{\text{Keterangan}}\\ 1.& \text{Penjumlahan}& \begin{array}{rl}& \text{Jumlahkan dua bilangan kompleks}\\ & \text{dengan cara bagian riil dan tidak riil}\\ & \text{(khayal murni) dilakukan secara}\\ & \text{terpisah}\\ & \mathbf{\text{Misalkan}}\\ & {\text{z}}_1=a+ib,\text{dan}{\text{z}}_2=c+id,\text{maka}\\ & {\text{z}}_1+{\text{z}}_2=(a+c)+(b+d)i\end{array}\\ 2.& \text{Pengurangan}& \begin{array}{rl}& \text{Pengurangan dua bilangan kompleks}\\ & \text{dengan cara bagian riil dan tidak riil}\\ & \text{(khayal murni) dilakukan secara}\\ & \text{terpisah}\\ & \mathbf{\text{Misalkan}}\\ & {\text{z}}_1=a+ib,\text{dan}{\text{z}}_2=c+id,\text{maka}\\ & {\text{z}}_1-{\text{z}}_2=(a-c)+(b-d)i\end{array}\\ 3.& \text{Perkalian}& \begin{array}{rl}& \text{Cara mengalikan dua bilangan }\\ & \text{kompleks yaitu lakukan dengan cara }\\ & \text{seperti binomial biasa dan ganti}\\ & i^2\text{dengan}\sqrt{-1}\\ & \text{Misalkan}\\ & {\text{z}}_1=a+ib,\text{dan}{\text{z}}_2=c+id,\text{maka}\\ & {\text{z}}_1\times {\text{z}}_2=(a+ib)\times (c+id)\\ & =ac+adi+bci+bd{i}^2\\ & =ac+bd{i}^2+(ad+bc)i\\ & =ac-bd+(ad+bc)i\\ & \mathbf{\text{Untuk perkalian dengan skalar}}\\ & \text{kalikan masing-masing bagian dengan}\\ & \text{skalarnya saja}\end{array}\\ 4.& \text{Pembagian}& \begin{array}{rl}& \text{cara membagi dua bilangan kompleks}\\ & \text{kalikan pembilang dan penyebut dengan}\\ & \text{sekawan dari penyebutnya serta ganti}\\ & \text{hasilnya yang mengandung}{i}^2\text{dengan}\\ & -1\\ & \text{Misalkan diketahui penyebutnya berupa}\\ & a+ib,\text{maka sekawannya adalah}:a-ib\\ & \text{atau}\\ & {\displaystyle \frac{{\text{z}}_1}{{\text{z}}_2}={\text{z}}_1{\text{z}}_2^{-1}\text{dengan}{\text{z}}_2\ne 0}\\ & \text{Jika}{\text{z}}_2=a+ib,\text{maka}\\ & {\text{z}}_2^{-1}={\displaystyle \frac{a}{a^2+b^2}-i\frac{b}{a^2+b^2}}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui bahwa}{\text{z}}_1=2+3i\text{dan}\\ & {\text{z}}_2=6-9i,\text{tentukan}\\ & \text{a}.{\text{z}}_1+{\text{z}}_2\text{c}.{\text{z}}_1\times {\text{z}}_2\\ & \text{b}.{\text{z}}_1-{\text{z}}_2\text{d}.{\text{z}}_1:{\text{z}}_2\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cl|}\hline \mathbf{\text{No}}& \mathbf{\text{Uraian jawaban}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \{\begin{array}{c}{\text{z}}_1=2+3i\\ {\text{z}}_2=6-9i\end{array}\end{array}\\ 1.& \begin{array}{rl}{\text{z}}_1+{\text{z}}_2& =(2+3i)+(6-9i)\\ & =(2+6)+(3-9)i\\ & =8-6i\end{array}\\ 2.& \begin{array}{rl}{\text{z}}_1-{\text{z}}_2& =(2+3i)-(6-9i)\\ & =(2-6)+(3+9)i\\ & =-8+12i\end{array}\\ 3.& \begin{array}{rl}{\text{z}}_1\times {\text{z}}_2& =(2+3i)\times (6-9i)\\ & =(2\times 6-3\times (-9))+(2\times (-9)+3\times 6)i\\ & =(12+27)+(-18+18)i\\ & =39\end{array}\\ 4.& \begin{array}{rl}{\text{z}}_1:{\text{z}}_2& ={\displaystyle \frac{{\text{z}}_1}{{\text{z}}_2}}\\ & ={\displaystyle \frac{2+3i}{6-9i}\times \frac{6+9i}{6+9i}}\\ & ={\displaystyle \frac{12+18i+18i+27i^2}{6^2+9^2}}\\ & ={\displaystyle \frac{12-27+(18+18)i}{36+81}}\\ & ={\displaystyle \frac{-15}{117}+\frac{36}{117}i=-{\displaystyle \frac{5}{39}+\frac{4}{13}i}}\\ \mathbf{\text{atau}}& \text{Anda dapat menggunakan invers}\\ & \text{hasilnyapun akan sama dengan yang}\\ & \text{di atas, yaitu}\\ & {\text{z}}_2^{-1}={\displaystyle \frac{6+9i}{6^2+9^2},\text{maka}}\\ & ={\text{z}}_1\times {\text{z}}_2^{-1}\\ & =(2+3i)\times {\displaystyle \frac{6+9i}{6^2+9^2}}\\ & ={\displaystyle \frac{12+18i+18i+27i^2}{36+81}}\\ & ={\displaystyle \frac{12-27+(18+18)i}{117}}\\ & ={\displaystyle \frac{-15}{117}+\frac{36}{117}i=-\frac{5}{39}+\frac{4}{13}i}\end{array}\\ \hline\end{array}\end{array}$.