# [Solving Complex Numbers] Division legal?

• June 1st 2010, 08:42 PM
Cthul
[Solving Complex Numbers] Division legal?
This is what I have done.
$(-4x+i)+(2-3yi)=6-2i$
Group real numbers and imaginary numbers
$(-4x+2)+(i-3yi)=6-2i$
$\Rightarrow$ Solve for real number
$-4x+2=6$

$-4x=4$

$x=-1$

$\Rightarrow$ Solve for imaginary number
$i-3yi=-2i$

$3yi=3i$

.. Can someone verify if it's legal to divide both sides by $3i$?
• June 1st 2010, 09:02 PM
pickslides
You can divide both sides by $3i$ but $y=1$ by inspection
• June 1st 2010, 09:05 PM
Cthul
• June 1st 2010, 09:08 PM
Prove It
Quote:

Originally Posted by Cthul
This is what I have done.
$(-4x+i)+(2-3yi)=6-2i$
Group real numbers and imaginary numbers
$(-4x+2)+(i-3yi)=6-2i$
$\Rightarrow$ Solve for real number
$-4x+2=6$

$-4x=4$

$x=-1$

$\Rightarrow$ Solve for imaginary number
$i-3yi=-2i$

$3yi=3i$

.. Can someone verify if it's legal to divide both sides by $3i$?

Yes, the division by $i$ is legal. However, these sorts of problems are easiest if you realise that for two complex expressions to be equal, their real parts have to be equal and their imaginary parts have to be equal.

So if you write each side in terms of its real and imaginary parts, then you can equate real and imaginary parts.

$(-4x+i)+(2-3yi)=6-2i$
$-4x + 2 + (1 - 3y)i = 6 - 2i$
$-4x + 2 = 6$ and $1 - 3y = -2$.