Equality of Complex Numbers

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Question 1 of 7

1. Question

Solve for $x$ and $y$ .

$5x-12i=9-4yi$

1. $x=9$ and $y=3$
2. $x=9/5$ and $y=3$
3. $x=3$ and $y=9/5$
4. $x=3$ and $y=9$

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Imaginary equations in the form

$a+bi=0$ include a real part

$a$ and an imaginary part

$bi$ .

To solve for

$x$ and

$y$ in

$5x-12i=9-4yi$ , set the real parts equal to each other and set the imaginary parts equal to each other.

Solve for

$x$ .

$color(red)(5x)+(-12)i=$	$color(red)(9)+(-4y)i$	The real parts are $5x$ and $9$ .
$color(red)(5x)=$	$color(red)(9)$	Set the real parts equal to each other.
$(5x)/5=$	$9/5$	Divide both sides by $5$ .
$x=$	$9/5$

Solve for

$y$ .

$5x color(blue)(-12i)$	$=9color(blue)(-4yi)$	The imaginary parts are $-12i$ and $-4yi$ .
$color(blue)(-12)=$	$color(blue)(-4y)$	Set the imaginary parts equal to each other.
$(-12)/(-4)=$	$(-4y)/(-4)$	Divide both sides by $-4$ .
$3=$	$y$	Reverse the equation.
$y=$	$3$

$x=9/5$ and

$y=3$

Question 2 of 7

2. Question

Solve for $x$ and $y$ .

$4x-1+(5-2y)i=3x+2+(y-1)i$

1. $x=3$ and $y=3$
2. $x=3$ and $y=2$
3. $x=3$ and $y=6$
4. $x=2$ and $y=3$

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Imaginary equations in the form

$a+bi=0$ include a real part

$a$ and an imaginary part

$bi$ .

To solve for

$x$ and

$y$ in

$4x-1+(5-2y)i=3x+2+(y-1)i$ , set the real parts equal to each other and set the imaginary parts equal to each other.

Solve for

$x$ .

$color(red)(4x-1)+(5-2y)i=$	$color(red)(3x+2)+(y-1)i$	The real parts are $4x-1$ and $3x+2$ .
$color(red)(4x-1)=$	$color(red)(3x+2)$	Set the real parts equal to each other.
$(4-3)x=$	$(2+1)$	Combine like terms.
$x=$	$3$

Solve for

$y$ .

$color(blue)((5-2y)i)$	$=color(blue)((y-1)i)$	The imaginary parts are $(5-2y)i$ and $(y-1)i$ .
$color(blue)((5-2y))=$	$color(blue)((y-1))$	Set the imaginary parts equal to each other.
$(-2-1)y=$	$(-1-5)$	Combine like terms.
$(-3y)/(-3)=$	$(-6)/(-3)$	Divide both sides by $-3$ .
$y=$	$2$

$x=3$ and

$y=2$

Question 3 of 7

3. Question

Solve for $x$ and $y$ .

$11+2x-4i=7x+5yi-4+6i$

1. $x=3$ and $y=-2$
2. $x=-3$ and $y=-2$
3. $x=-3$ and $y=2$
4. $x=3$ and $y=2$

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Imaginary equations in the form

$a+bi=0$ include a real part

$a$ and an imaginary part

$bi$ .

To solve for

$x$ and

$y$ in

$11+2x-4i=7x+5yi-4+6i$ , set the real parts equal to each other and set the imaginary parts equal to each other.

Solve for

$x$ .

$color(red)(11+2x)-4i=$	$color(red)(7x-4)+(5y+6)i$	The real parts are $11+2x$ and $7x-4$ .
$color(red)(11+2x)=$	$color(red)(7x-4)$	Set the real parts equal to each other.
$(2-7)x=$	$(-4-11)$	Combine like terms.
$(-5x)/(-5)=$	$(-15)/(-5)$	Divide both sides by $-5$ .
$x=$	$3$

Solve for

$y$ .

$11+2x color(blue)(-4i)$	$=7x-4 color(blue)(+(5y+6)i)$	The imaginary parts are $-4i$ and $(5y+6)i$ .
$color(blue)((-4))=$	$color(blue)((5y+6))$	Set the imaginary parts equal to each other.
$(-5)y=$	$(6+4)$	Combine like terms.
$(-5y)/(-5)=$	$(10)/(-5)$	Divide both sides by $-3$ .
$y=$	$-2$

$x=3$ and

$y=-2$

Question 4 of 7

4. Question

Solve for $x$ and $y$ .

$(x+4i)(1-i)=9+yi$

1. $x=-5$ and $y=1$
2. $x=5$ and $y=1$
3. $x=5$ and $y=-1$
4. $x=-5$ and $y=-1$

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Imaginary equations in the form

$a+bi=0$ include a real part

$a$ and an imaginary part

$bi$ .

To solve for

$x$ and

$y$ in

$(x+4i)(1-i)=9+yi$ , expand, simplify, and then set the real parts equal to each other and set the imaginary parts equal to each other.

Expand and simplify the left-hand side of the equation.

$color(blue)((x+4i)(1-i))=$	$9+yi$	Expand the left-hand side of the equation.
$x-x\i+4i-4i^2=$	$9+yi$	Simplify. Remember that $i^2=-1$ .
$x-x\i+4i+4=$	$9+yi$	Group the real parts together $x$ and $4$ .
$(x+4)+i(-x+4)=$	$9+yi$	Group the imaginary parts together $-x\i$ and $4i$ .

Solve for

$x$ .

$color(red)((x+4))+i(-x+4)=$	$color(red)(9)+yi$	The real parts are $x+4$ and $9$ .
$color(red)(x+4)=$	$color(red)(9)$	Set the real parts equal to each other.
$x+4-4=$	$9-4$	Subtract $4$ from each side.
$x=$	$5$

Solve for

$y$ .

$(x+4)color(blue)(+i(-x+4))=$	$9color(blue)(+yi)$	The imaginary parts are $i(-x+4)$ and $yi$ .
$color(blue)(i(-x+4))=$	$color(blue)(yi)$	Set the imaginary parts equal to each other.
$(i(-x+4))/i=$	$(yi)/i$	Divide both sides by $i$ .
$-x+4=$	$y$	Sub in $x=5$
$-(5)+4=$	$y$	Simplify
$-1=$	$y$	Reverse the sides
$y=$	$-1$

$x=5$ and

$y=-1$

Question 5 of 7

5. Question

Solve for $x$ and $y$ .

$(2+2i)(x+yi)=4+12i$

1. $x=-4$ and $y=2$
2. $x=-4$ and $y=-2$
3. $x=4$ and $y=2$
4. $x=4$ and $y=-2$

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Imaginary equations in the form

$a+bi=0$ include a real part

$a$ and an imaginary part

$bi$ .

To solve for

$x$ and

$y$ in

$(2+2i)(x+yi)=4+12i$ , expand, simplify, and then set the real parts equal to each other and set the imaginary parts equal to each other.

Expand and simplify the left-hand side of the equation.

$color(blue)((2+2i)(x+yi))=$	$4+12i$	Expand the left-hand side of the equation.
$2x+2yi+2xi+2yi^2=$	$4+12i$	Simplify. Remember that $i^2=-1$ .
$(2x-2y)+2yi+2xi=$	$4+12i$	Group the real parts together.
$(2x-2y)+(2x+2y)i=$	$4+12i$	Group the imaginary parts together.

Solve for

$x$ .

$color(red)(2x-2y)+(2x+2y)i=$	$color(red)(4)+12i$	The real parts are $2x-2y$ and $4$ .
$color(red)(2x-2y)=$	$color(red)(4)$	Set the real parts equal to each other.
$2x-2y=$	$4$	Set as Equation 1.

Solve for

$y$ .

$2x-2y color(blue)(+(2x+2y)i)=$	$4color(blue)(+12i)$	The imaginary parts are $i(2x+2y)$ and $12i$ .
$color(blue)(i(2x+2y))=$	$color(blue)(12i)$	Set the imaginary parts equal to each other.
$(i(2x+2y))/i=$	$(12i)/i$	Divide both sides by $i$ .
$2x+2y=$	$12$	Set as Equation 2.

Subtracting Equation 1 and Equation 2.

$color(blue)(2x-2y)-color(red)(2x+2y)=$	$color(blue)(4)-color(red)(12)$	Subtract Equation 1 and Equation 2.
$4y=$	$8$	Combine like terms
$(4y)/4=$	$8/4$	Divide both sides by $4$ .
$y=$	$2$

Substitute

$y=2$ to Equation 2.

$2x+2(2)=$	$12$	Substitute y = 2 to Equation 2.
$2x=$	$12-4$	Combine like terms
$(2x)/2=$	$8/2$	Divide both sides by $2$ .
$x=$	$4$

$x=4$ and

$y=2$

Question 6 of 7

6. Question

Solve for $x$ and $y$ .

$(2x+yi)(1-i)=2i$

1. $x=1$ and $y=(-1)/2$
2. $x=(-1)/2$ and $y=1$
3. $x=2$ and $y=-1$
4. $x=1$ and $y=2$

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Imaginary equations in the form

$a+bi=0$ include a real part

$a$ and an imaginary part

$bi$ .

To solve for

$x$ and

$y$ in

$(2x+yi)(1-i)=2i$ , expand, simplify, and then set the real parts equal to each other and set the imaginary parts equal to each other.

Expand and simplify the left-hand side of the equation.

$color(blue)((2x+yi)(1-i))=$	$2i$	Expand the left-hand side of the equation.
$2x-2x\i+yi-yi^2=$	$2i$	Simplify. Remember that $i^2=-1$ .
$2x-2x\i+yi+y=$	$2i$	Group the real parts together.
$(2x+y)+i(-2x+y)=$	$2i$	Group the imaginary parts together.

Solve for

$x$ .

$color(red)(2x+y)+(-2x+y)i=$	$2i$	The real parts are $2x+y$ and $0$ .
$color(red)(2x+y)=$	$0$	Set the real parts equal to each other.
$2x+y=$	$0$	Set as Equation 1.

Solve for

$y$ .

$2x+y color(blue)((-2x+y)i)=$	$color(blue)(2i)$	The imaginary parts are $i(-2x+y)$ and $2i$ .
$color(blue)(i(-2x+y))=$	$color(blue)(2i)$	Set the imaginary parts equal to each other.
$(i(-2x+y))/i=$	$(2i)/i$	Divide both sides by $i$ .
$-2x+y=$	$2$	Set as Equation 2.

Adding Equation 1 and Equation 2.

$color(blue)(2x+y)+color(red)(-2x+y)=$	$color(blue)(0)+color(red)(2)$	Subtract Equation 1 and Equation 2.
$2y=$	$2$	Combine like terms
$(2y)/2=$	$2/2$	Divide both sides by $4$ .
$y=$	$1$

Substitute

$y=1$ to Equation 1.

$2x+1=$	$0$	Substitute y = 2 to Equation 2.
$2x=$	$-1$	Combine like terms
$(2x)/2=$	$(-1)/2$	Divide both sides by $2$ .
$x=$	$(-1)/2$

$x=(-1)/2$ and

$y=1$

Question 7 of 7

7. Question

Solve for $x$ and $y$ .

$(x-yi)(3+i)=i$

1. $x=1/10$ and $y=(-3)/10$
2. $x=-1$ and $y=3$
3. $x=1$ and $y=-3$
4. $x=(-1)/10$ and $y=3/10$

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Imaginary equations in the form

$a+bi=0$ include a real part

$a$ and an imaginary part

$bi$ .

To solve for

$x$ and

$y$ in

$(x-yi)(3+i)=i$ , expand, simplify, and then set the real parts equal to each other and set the imaginary parts equal to each other.

Expand and simplify the left-hand side of the equation.

$color(blue)((x-yi)(3+i)=i)=$	$i$	Expand the left-hand side of the equation.
$3x+x\i-3yi-yi^2=$	$i$	Simplify. Remember that $i^2=-1$ .
$(3x+y)+x\i-3yi=$	$i$	Group the real parts together.
$(3x+y)+i(x-3y)=$	$i$	Group the imaginary parts together.

Solve for

$x$ .

$color(red)(3x+y)+(x-3y)i=$	$i$	The real parts are $3x+y$ and $0$ .
$color(red)(3x+y)=$	$0$	Set the real parts equal to each other.
$3x+y=$	$0$	Set as Equation 1.

Solve for

$y$ .

$3x + y color(blue)(+(x-3y)i)=$	$color(blue)(i)$	The imaginary parts are $(x-3y)i$ and $i$ .
$color(blue)(i(x-3y))=$	$color(blue)(i)$	Set the imaginary parts equal to each other.
$(i(x-3y))/i=$	$i/i$	Divide both sides by $i$ .
$x-3y=$	$1$	Set as Equation 2.

Solving for

$x$ .

$(3x+y)times3=$	$0times3$	Multiply the whole equation by 3.
$9x+3y=$	$0$	Set as Equation 3.
$color(blue)(x-3y)+color(red)(9x+3y)=$	$color(blue)(1)+color(red)(0)$	Add Equation 2 and Equation 3.
$(1+9)x=$	$1$	Combine like terms.
$(10x)/10=$	$1/10$	Divide both sides by $10$ .
$x=$	$1/10$

Solving for

$y$ .

$3(1/10)+y=$	$0$	Substitute $x = 1/10$ to Equation 1.
$y=$	$(-3)/10$	Combine like terms.
$y=$	$(-3)/10$

$x=1/10$ and

$y=(-3)/10$

Quizzes

Simplify Roots of Negative Numbers 1
Simplify Roots of Negative Numbers 2
Powers of the Imaginary Unit 1
Powers of the Imaginary Unit 2
Solve Quadratic Equations with Complex Solutions 1
Solve Quadratic Equations with Complex Solutions 2
Equality of Complex Numbers
Add and Subtract Complex Numbers 1
Add and Subtract Complex Numbers 2
Multiply Complex Numbers 1
Multiply Complex Numbers 2
Divide Complex Numbers
Complex Numbers – Product of Linear Factors 1
Complex Numbers – Product of Linear Factors 2
Mixed Operations with Complex Numbers