1

How do you solve equations involving $z = a + bi$ and imaginary units?

The one I am looking at right now:

$$\frac{z-2}{z+1} = 3i$$

If you could help me with this one, I think I can do the rest by myself.

Pauly B
  • 5,272
Oscar
  • 11

2 Answers2

4

Rewrite as $z-2=3iz+3i$ and isolate $z$ as usual. We get $$z=\frac{2+3i}{1-3i}.$$ You may be expected to change the form of the answer by multiplying top and bottom by $1+3i$.

André Nicolas
  • 507,029
  • Oh, so it's just like we normally do? That's convenient! Thanks. – Oscar Sep 05 '14 at 16:55
  • 1
    You are welcome. It is a disguised linear equation. When you do the suggested complex number manipulation, you should end up with $-\frac{7}{10}+\frac{9}{10}i$. – André Nicolas Sep 05 '14 at 17:02
1

Set $z=a+ib$ where $a,b$ are real

So, we have $a+ib-2=3i(a+1+i)$

$$a-2+ib=-3+i\cdot3(a+1)$$

Now equate the real & the imaginary parts

lab bhattacharjee
  • 274,582