Complex Equations

If two complex numbers are equal then the real and imaginary parts are also equal. We call this equating like parts. 

Example 1

If $a+6i=3+6i$a+6i=3+6i, then $a=3$a=3

If $8-bi=8+7i$8−bi=8+7i, then $b=-7$b=−7

 

We can use this process to solve algebraic problems involving complex numbers.

Example 2

Find $x,y$x,y if $\left(3+2i\right)^2-3\left(x+iy\right)=x+iy$(3+2i)2−3(x+iy)=x+iy

	$\left(3+2i\right)^2-3\left(x+iy\right)$(3+2i)2−3(x+iy)	$=$=	$x+iy$x+iy
	$9+12i-4-3x-3yi$9+12i−4−3x−3yi	$=$=	$x+iy$x+iy
	$5-3x+12i-3yi$5−3x+12i−3yi	$=$=	$x+iy$x+iy
	$(5-3x)+i(12-3y)$(5−3x)+i(12−3y)	$=$=	$x+iy$x+iy
	Therefore:
	$5-3x$5−3x	$=$=	$x$x
	$5$5	$=$=	$4x$4x
	$x$x	$=$=	$\frac{5}{4}$54​
	and
	$12-3y$12−3y	$=$=	$y$y
	$12$12	$=$=	$4y$4y
	$y$y	$=$=	$3$3

 

 

 

Example 3

Find $x,y$x,y if $\frac{x}{1-i}+\frac{y}{4+3i}=2-4i$x1−i​+y4+3i​=2−4i

$\frac{x}{1-i}+\frac{y}{4+3i}$x1−i​+y4+3i​	$=$=	$2-4i$2−4i
$\frac{x\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}+\frac{y\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}$x(1+i)(1−i)(1+i)​+y(4−3i)(4+3i)(4−3i)​	$=$=	$2-4i$2−4i
$\frac{x+xi}{1+1}+\frac{4y-3yi}{16+9}$x+xi1+1​+4y−3yi16+9​	$=$=	$2-4i$2−4i
$\frac{x+xi}{2}+\frac{4y-3yi}{25}$x+xi2​+4y−3yi25​	$=$=	$2-4i$2−4i
$\frac{25x+25xi}{50}+\frac{8y-6yi}{50}$25x+25xi50​+8y−6yi50​	$=$=	$2-4i$2−4i
$25x+25xi+8y-6yi$25x+25xi+8y−6yi	$=$=	$100-200i$100−200i
$\left(25x+8y\right)+i\left(25x-6y\right)$(25x+8y)+i(25x−6y)	$=$=	$100-200i$100−200i
Therefore
$25x+8y$25x+8y	$=$=	$100$100 (1)
$25x-6y$25x−6y	$=$=	$-200$−200 (2)
By equating like parts we can then get simultaneous equations to solve
(1)-(2)                                          $14y$14y	$=$=	$300$300
$y$y	$=$=	$\frac{300}{14}$30014​
Sub $y=\frac{300}{14}$y=30014​ into (1)                         $25x+8\times\frac{300}{14}$25x+8×30014​	$=$=	$100$100
$25x$25x	$=$=	$\frac{-500}{7}$−5007​
$x$x	$=$=	$-\frac{20}{7}$−207​

More Worked Examples

QUESTION 1

QUESTION 2