Equality of Complex Numbers

Question

Find exact x, y such that 

 

a) (x+2i)(y-i)= -4-7i 

 

b) (x+i)(3-iy)= 1+13i 

Answer 1

For each part, you want to multiply the left side using FOIL.  Then equate coefficients of real and imaginary terms.

 

a)

 

xy + 2yi - xi + 2 = -4 - 7i

 

(xy + 2) + (2y - x)i = -4 - 7i

 

 

Real:  xy + 2 = -4

imag: 2y - x = -7

 

These equations that represent real coefficients and imaginary coefficients are your system of equations.  Solve them for x and y using substitution/elimination methods.

 

 

Do the same for part B.

Answer 2

a) (x+2i)(y-i)=-4-7i

 

xy+2yi-xi+2=-4-7i

xy+2+2yi-xi=-4-7i

xy+2=-4

(2y-x)i=-7i

2y-x=-7 along with xy+2=-4

using the first equation...

x=2y+7

substituting this into the second equation...

y(2y+7)+2=-4

2y^2+7y+6=0

factor

(2y+3)(y+2)=0

2y+3=0

2y=-3

y=-3/2

y+2=0

y=-2

y=-3/2 and y=-2

using y=-3/2, find x

2y-x=-7

2(-3/2)-x=-7

-3-x=-7

x=4

x=4 and y=-3/2

using y=-2, find x

xy+2=-4

-2x+2=-4

-2x=-6

x=3

x=3 and y=-2

 

b) (x+i)(3-iy)=1+13i

 

3x+3i-xyi+y=1+13i

3x+y+(3-xy)i=1+13i

3x+y=1

3-xy=13

y=1-3x from 3x+y=1

3-x(1-3x)=13

3-x+3x^2=13

3x^2-x-10=0

(3x+5)(x-2)=0

can you finish up from here ?