Soal dan Pembahasan Bilangan Kompleks 2. Operasi Aljabar dan Invers Bilangan Kompleks

Soal No. 1

Jika $z_1=2-5i$ dan $z_2=3+4i$ maka nilai dari $3z_1+2z_2$ = …. A. $12-7i$
B. $12+7i$
C. $7+12i$
D. $7-12i$
E. $19i$

Penyelesaian: Lihat/Tutup

$\begin{align}3z_1+2z_2 &= 3(2-5i)+2(3+4i) \\ &= 6-15i+6+8i \\ &= 12-7i \end{align}$
Jawaban: A

Soal No. 2

Jika $z_1=2+\sqrt{3}i$ dan $z_2=3-\frac{1}{\sqrt{3}}i$ maka nilai dari $3z_1-2z_2$ = ….
A. $12-\frac{11}{3}\sqrt{3}i$
B. $12+\frac{11}{3}\sqrt{3}i$
C. $-\frac{11}{3}\sqrt{3}i$
D. $\frac{11}{3}\sqrt{3}i$
E. 12

Penyelesaian: Lihat/Tutup

$\begin{align}3z_1-2z_2 &= 3(2+\sqrt{3}i)-2\left( 3-\frac{1}{\sqrt{3}}i \right) \\ &= 6+3\sqrt{3}i-6+\frac{2}{\sqrt{3}}i \\ &= \frac{3\sqrt{3}i}{1}+\frac{2i}{\sqrt{3}} \\ &= \frac{9i+2i}{\sqrt{3}} \\ &= \frac{11i}{\sqrt{3}} \\ &= \frac{11}{3}\sqrt{3}i \end{align}$
Jawaban: D

Soal No. 3

Jika $z_1=3+2i$ dan $z_2=3-2i$ maka $z_1^2-z_2^2$ = ….
A. 5
B. 10
C. $12i$
D. $24i$
E. $10+24i$

Penyelesaian: Lihat/Tutup

$\begin{align}z_1^2-z_2^2 &= (z_1+z_2)(z_1-z_2) \\ &= (3+2i+3-2i)(3+2i-3+2i) \\ &= 6(4i) \\ &= 24i \end{align}$
Jawaban: D

Soal No. 4

Jika $z_1=2-i$ dan $z_2=-3+i$ maka $z_1^3+z_2^3$ = ….
A. $-16-15i$
B. $-16+15i$
C. $16-15i$
D. $16+15i$
E. $25i$

Penyelesaian: Lihat/Tutup

$\begin{align}z_1^3 &= (2-i)^3 \\ &= (2-i)^2(2-i) \\ &= (4-4i+i^2)(2-i) \\ &= (4-4i-1)(2-i) \\ &= (3-4i)(2-i) \\ &= 6-3i-8i+4i^2 \\ &= 6-11i+4(-1) \\ z_1^3 &= 2-11i \end{align}$
$\begin{align}z_2^3 &= (-3+i)^3 \\ &= (-3+i)^2(-3+i) \\ &= (9-6i+i^2)(-3+i) \\ &= (9-6i-1)(-3+i) \\ &= (8-6i)(-3+i) \\ &= -24+8i+18i-6i^2 \\ &= -24+26i-6(-1) \\ z_1^3 &= -18+26i \end{align}$
$\begin{align}z_1^3+z_2^3 &= 2-11i+(-18+26i) \\ &= -16+15i \end{align}$
Jawaban: B

Soal No. 5

Jika bilangan kompleks $z_1=(1,1)$ dan $z_1.z_2=1$ maka $z_2$ = ….
A. $\left( -\frac{1}{2},-\frac{1}{2} \right)$
B. $\left( -\frac{1}{2},\frac{1}{2} \right)$
C. $\left( \frac{1}{2},-\frac{1}{2} \right)$
D. $\left( \frac{1}{2},\frac{1}{2} \right)$
E. $(-1,-1)$

Penyelesaian: Lihat/Tutup

$z_1.z_2=1$ maka:
$\begin{align}z_2 &= z_1^{-1} \\ &= \frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i \\ &= \frac{1}{1^2+1^2}-\frac{1}{1^2+1^2}i \\ &= \frac{1}{2}-\frac{1}{2}i \\ z_2 &= \left( \frac{1}{2},-\frac{1}{2} \right) \end{align}$
Jawaban: C

Soal No. 6

Bagian real dan bagian imaginer bilangan kompleks $\frac{3+2i}{1+i}$ berturut-turut adalah ….
A. $-\frac{5}{2}$ dan $-\frac{1}{2}$
B. $\frac{5}{2}$ dan $-\frac{1}{2}$
C. $-\frac{5}{2}$ dan $\frac{1}{2}$
D. $\frac{5}{2}$ dan $\frac{1}{2}$
E. 5 dan 1

Penyelesaian: Lihat/Tutup

$\begin{align}z &= \frac{3+2i}{1+i}\times \frac{1-i}{1-i} \\ &= \frac{3-3i+2i-2i^2}{1-i+i-i^2} \\ &= \frac{3-i-2(-1)}{1-(-1)} \\ &= \frac{5-i}{2} \\ z &= \frac{5}{2}-\frac{1}{2}i \end{align}$
$\operatorname{Re}(z)=\frac{5}{2}$ dan $\operatorname{Im}(z)=-\frac{1}{2}$
Jawaban: B

Soal No. 7

Invers dari bilangan kompleks $\frac{3}{i}$ adalah ….
A. $-\frac{2}{3}i$
B. $-\frac{1}{3}i$
C. 1
D. $\frac{1}{3}i$
E. $\frac{2}{3}i$

Penyelesaian: Lihat/Tutup

$\begin{align}z &= \frac{3}{i} \\ &= \frac{3}{i}\times \frac{i}{i} \\ &= \frac{3i}{i^2} \\ &= \frac{3i}{-1} \\ z &= -3i \\ z &= (0,-3) \end{align}$
$\begin{align}z^{-1} &= \frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i \\ &= \frac{0}{{{0}^{2}}+{{(-3)}^{2}}}-\frac{-3}{{{0}^{2}}+{{(-3)}^{2}}}i \\ &= 0+\frac{3}{9}i \\ z^{-1} &= \frac{1}{3}i \end{align}$
Jawaban: D

Soal No. 8

Bilangan real $x$ dan $y$ berturut-turut yang memenuhi $(2+3i)x+(2-3i)y=2-i$ adalah ….
A. $\frac{1}{4}$ dan $\frac{3}{4}$
B. $\frac{1}{3}$ dan $\frac{2}{3}$
C. $\frac{1}{2}$ dan $\frac{1}{2}$
D. $-\frac{1}{4}$ dan $\frac{5}{4}$
E. $-\frac{1}{3}$ dan $\frac{4}{3}$

Penyelesaian: Lihat/Tutup

$\begin{align}(2+3i)x+(2-3i)y &= 2-i \\ 2x+3xi+2y-3yi &= 2-i \\ (2x+2y)+(3x-3y)i &= 2-i \end{align}$
$\begin{align}2x+2y &= 2 \\ x+y &= 1 \\ x &= 1-y \end{align}$
$\begin{align}3x-3y &= -1 \\ 3(1-y)-3y &= -1 \\ 3-3y-3y &= -1 \\ -6y &= -4 \\ y &= \frac{-4}{-6} \\ y &= \frac{2}{3} \end{align}$
$\begin{align}x &= 1-y \\ &= 1-\frac{2}{3} \\ x &= \frac{1}{3} \end{align}$
Jadi, bilangan real $x$ dan $y$ yang memenuhi adalah $\frac{1}{3}$ dan $\frac{2}{3}$.
Jawaban: B

Soal No. 9

Jika bilangan real $x$ dan $y$ memenuhi $\frac{x-2}{2+i}+\frac{y-2}{2-i}=4$ maka $(x+yi)^2+(x-yi)^2$ = ….
A. -81
B. -80
C. 0
D. 80
E. 81

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{x-2}{2+i}+\frac{y-2}{2-i} &= 4 \\ \frac{x-2}{2+i}\times \frac{2-i}{2-i}+\frac{y-2}{2-i}\times \frac{2+i}{2+i} &= 4 \\ \frac{2x-xi-4+2i}{4-2i+2i-i^2}+\frac{2y+yi-4-2i}{4+2i-2i-i^2} &= 4 \\ \frac{2x+2y-8+(y-x)i}{4-(-1)} &= 4 \\ \frac{2x+2y-8+(y-x)i}{5} &= 4 \\ 2x+2y-8+(y-x)i &= 20 \end{align}$
$\begin{align}2x+2y-8 &= 20 \\ 2x+2y &= 28 \\ x+y &= 14 \\ x &= 14-y \end{align}$
$\begin{align}y-x &= 0 \\ y-(14-y) &= 0 \\ y-14+y &= 0 \\ 2y &= 14 \\ y &= 7 \end{align}$
$\begin{align}x &= 14-y \\ &= 14-7 \\ x &= 7 \end{align}$
${{(x+yi)}^{2}}+{{(x-yi)}^{2}}$
= ${{(7+7i)}^{2}}+{{(7-7i)}^{2}}$
= $49+98i+49i^2+49-98i+49i^2$
= $98+98i^2$
= $98+98(-1)$
= 0
Jawaban: C

Soal No. 10

Bilangan kompleks $(1+i)^{2022}$ = ….
A. $-2^{2022}$
B. $-2^{1011}i$
C. $i$
D. $2^{1011}i$
E. $2^{2022}i$

Penyelesaian: Lihat/Tutup

$\begin{align}(1+i)^{2022} &= \left( (1+i)^2 \right)^{1011} \\ &= (1+2i+i^2)^{1011} \\ &= (1+2i+-1)^{1011} \\ &= (2i)^{1011} \\ &= 2^{1011}.i^{1011} \\ &= 2^{1011}.i^{1010}.i \\ &= 2^{1011}.(i^2)^{505}.i \\ &= 2^{1011}.(-1)^{505}.i \\ (1+i)^{2022} &= -2^{1011}i \end{align}$
Jawaban: B

Soal No. 11

Misal $z=a+bi$ adalah bilangan kompleks dan $z^2=i$ maka $a^2-b^2$ = ….
A. -2
B. -1
C. 0
D. 1
E. 2

Penyelesaian: Lihat/Tutup

$\begin{align}z^2 &= i \\ (a+bi)^2 &= i \\ a^2+2abi+b^2i^2 &= i \\ a^2+2abi+b^2(-1) &= i \\ a^2-b^2+2abi &= 0+i \end{align}$
$a^2-b^2=0$
Jawaban: C

Soal No. 12

Jika $z=-1-i$ maka $z^2+2z+2$ = ….
A. -2
B. -1
C. 0
D. 1
E. 2

Penyelesaian: Lihat/Tutup

$z^2+2z+2$
= $(-1-i)^2+2(-1-i)+2$
= $1+2i+i^2-2-2i+2$
= $1+i^2$
= $1-1$
= 0
Jawaban: C

Soal No. 13

Jika bilangan kompleks $z=\sqrt{3}+i$ maka $z^6$ = ….
A. -64
B. -32
C. 0
D. 32
E. 6

Penyelesaian: Lihat/Tutup

$z=\sqrt{3}+i$
$\begin{align}z^2 &= \left( \sqrt{3}+i \right)^2 \\ &= 3+2\sqrt{3}i+i^2 \\ z^2 &= 2+2\sqrt{3}i \end{align}$
$\begin{align}z^4 &= \left( z^2 \right)^2 \\ &= \left( 2+2\sqrt{3}i \right)^2 \\ &= 4+8\sqrt{3}i+12i^2 \\ z^4 &= -8+8\sqrt{3}i \end{align}$
$\begin{align}z^6 &= z^2.z^4 \\ &= \left( 2+2\sqrt{3}i \right)\left( -8+8\sqrt{3}i \right) \\ &= -16+16\sqrt{3}i-16\sqrt{3}i+48i^2 \\ &= -16+48(-1) \\ z^6 &= -64 \end{align}$
Jawaban: A

Soal No. 14

Persamaan kuadrat yang akar-akarnya $x_1=i$ dan $x_2=\frac{1}{1-i}$ adalah ….
A. $x^2-\frac{1}{2}x-\frac{1}{2}=0$
B. $x^2-\frac{3}{2}ix+\frac{1}{2}i=0$
C. $x^2-\frac{1}{2}x+\frac{1}{2}i+\frac{1}{2}=0$
D. $x^2-\left( \frac{1}{2}+\frac{3}{2}i \right)x+\frac{1}{2}i+\frac{1}{2}=0$
E. $x^2-\left( \frac{1}{2}+\frac{3}{2}i \right)x+\frac{1}{2}i-\frac{1}{2}=0$

Penyelesaian: Lihat/Tutup

$x_1=i$
$\begin{align}x_2 &= \frac{1}{1-i}\times \frac{1+i}{1+i} \\ &= \frac{1+i}{1-i^2} \\ &= \frac{1+i}{1-(-1)} \\ &= \frac{1+i}{2} \\ x_2 &= \frac{1}{2}+\frac{1}{2}i \end{align}$
$x_1+x_2=i+\frac{1}{2}+\frac{1}{2}i=\frac{1}{2}+\frac{3}{2}i$
$\begin{align}x_1\times x_2 &= i\left( \frac{1}{2}+\frac{1}{2}i \right) \\ &= \frac{1}{2}i+\frac{1}{2}i^2 \\ &= \frac{1}{2}i+\frac{1}{2}(-1) \\ x_1\times x_2 &= \frac{1}{2}i-\frac{1}{2} \end{align}$
Persamaan kuadrat yang akar-akarnya $x_1$ dan $x_2$ adalah:
$\begin{align}x^2-(x_1+x_2)x+x_1x_2 &= 0 \\ x^2-\left( \frac{1}{2}+\frac{3}{2}i \right)x+\frac{1}{2}i-\frac{1}{2} &= 0 \end{align}$
Jawaban: E

Soal No. 15

Jika bilangan kompleks $z_1=4-3i$, $z_2=2-2i$, $z_3=3+i$ dan $z_4=-3+2i$ maka $4\frac{z_1}{z_2}+5\frac{z_2}{z_3}+13\frac{z_3}{z_4}+25\frac{z_4}{z_1}$ = ….
A. $-2-13i$
B. $-2+13i$
C. $2-13i$
D. $2+13i$
E. $13+2i$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{z_1}{z_2} &= \frac{4-3i}{2-2i}\times \frac{2+2i}{2+2i} \\ &= \frac{8+8i-6i-6i^2}{4+4i-4i-4i^2} \\ &= \frac{8+2i-6(-1)}{4-4(-1)} \\ \frac{z_1}{z_2} &= \frac{14+2i}{8} \end{align}$
$\begin{align}\frac{z_2}{z_3} &= \frac{2-2i}{3+i}\times \frac{3-i}{3-i} \\ &= \frac{6-2i-6i+2i^2}{9-3i+3i-i^2} \\ &= \frac{6-8i+2(-1)}{9-(-1)} \\ \frac{z_2}{z_3} &= \frac{4-8i}{10} \end{align}$
$\begin{align}\frac{z_3}{z_4} &= \frac{3+i}{-3+2i}\times \frac{-3-2i}{-3-2i} \\ &= \frac{-9-6i-3i-2i^2}{9+6i-6i-4i^2} \\ &= \frac{-9-9i-2(-1)}{9-4(-1)} \\ \frac{z_3}{z_4} &= \frac{-7-9i}{13} \end{align}$
$\begin{align}\frac{z_4}{z_1} &= \frac{-3+2i}{4-3i}\times \frac{4+3i}{4+3i} \\ &= \frac{-12-9i+8i+6i^2}{16+12i-12i-9i^2} \\ &= \frac{-12-i+6(-1)}{16-9(-1)} \\ \frac{z_4}{z_1} &= \frac{-18-i}{25} \end{align}$
$4\frac{z_1}{z_2}+5\frac{z_2}{z_3}+13\frac{z_3}{z_4}+25\frac{z_4}{z_1}$
= $4.\frac{14+2i}{8}$ + $5.\frac{4-8i}{10}$ + $13.\frac{-7-9i}{13}$ + $25.\frac{-18-i}{25}$
= $\frac{14+2i}{2}$ + $\frac{4-8i}{2}$ + $-7-9i$ + $-18-i$
= $7+i$ + $2-4i$ + $-7-9i$ + $-18-i$
= $-16-13i$
Jawaban: A

Soal No. 16

Dua bilangan kompleks $5+2i$ dan $3+4i$ bila dikalikan hasilnya adalah …
A. $2+23i$
B. $5+26i$
C. $7+23i$
D. $7+26i$
E. $23+26i$

Penyelesaian: Lihat/Tutup

$\begin{align}(5+2i)(3+4i) &= 15+20i+6i+8i^2 \\ &= 15+26i+8(-1) \\ &= 7+26i \end{align}$
Jawaban: D

Soal No. 17

Ditentukan $(2+3i)z=2+i$. Jika $z$ bilangan kompleks, nilai $z$ = …
A. $\frac{1}{13}(7-4i)$
B. $\frac{1}{5}(7-4i)$
C. $\frac{1}{5}(7+4i)$
D. $\frac{1}{13}(7+4i)$
E. $\frac{1}{13}(1-4i)$

Penyelesaian: Lihat/Tutup

$\begin{align}(2+3i)z &= 2+i \\ z &= \frac{2+i}{2+3i}\times \frac{2-3i}{2-3i} \\ &= \frac{4-6i+2i-3i^2}{4-6i+6i-9i^2} \\ &= \frac{4-4i-3(-1)}{4-9(-1)} \\ z &= \frac{7-4i}{13} \\ z &= \frac{1}{13}(7+4i) \\ \end{align}$
Jawaban: D

Soal No. 18

Ditentukan $z_1=2+3i$ dan $z_2=1-3i$, maka bagian imajiner dari $\frac{z_1}{z_2}$ adalah …
A. $-\frac{9}{10}$
B. $-\frac{3}{8}$
C. $\frac{9}{10}$
D. $\frac{11}{10}$
E. $\frac{9}{8}$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{z_1}{z_2} &= \frac{2+3i}{1-3i} \\ &= \frac{2+3i}{1-3i}\times \frac{1+3i}{1+3i} \\ &= \frac{2+6i+3i+9i^2}{1+3i-3i-9i^2} \\ &= \frac{2+9i+9(-1)}{1-9(-1)} \\ &= \frac{-7+9i}{10} \\ \frac{z_1}{z_2} &= -\frac{7}{10}+\frac{9}{10}i \end{align}$
$\operatorname{Im}\left( \frac{z_1}{z_2} \right)=\frac{9}{10}$
Jawaban: C

Soal No. 19

Diketahui $z_1=2-\sqrt{2}i$ dan $z_2=4+4\sqrt{2}i$. Tentukan $z_1+z_2$.
A. $6+3\sqrt{2}i$
B. $4+\sqrt{2}i$
C. $6+4\sqrt{2}i$
D. $4-\sqrt{2}i$
E. $6-3\sqrt{2}i$

Penyelesaian: Lihat/Tutup

$\begin{align}z_1+z_2 &= 2-\sqrt{2}i+4+4\sqrt{2}i \\ &= 6+3\sqrt{2}i \end{align}$
Jawaban: A

Soal No. 20

Tentukan invers terhadap perkalian $z=1+2i$
A. $-\frac{1}{5}-\frac{2}{5}i$
B. $-\frac{1}{5}-\frac{3}{5}i$
C. $\frac{1}{5}+\frac{3}{5}i$
D. $\frac{1}{5}-\frac{2}{5}i$
E. $-\frac{1}{5}+\frac{2}{5}i$

Penyelesaian: Lihat/Tutup

$z=1+2i=(1,2)$ diperoleh $x=1$ dan $y=2$
$\begin{align}z^{-1} &= \frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i \\ &= \frac{1}{1^2+2^2}-\frac{2}{1^2+2^2}i \\ z^{-1} &= \frac{1}{5}-\frac{2}{5}i \end{align}$
Jawaban: D

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