+0

# complex numbers

0
6
3
+4

describe all solutions to zw-3w-2iw=4iz=-8+12i

Jan 28, 2024

#1
+14826
0

$$zw-3w-2iw=4iz=-8+12i$$

$$4iz=-8+12i\\ z=\frac{-8+12i}{4i}\\ \color{blue}z=3+2i$$

$$zw-3w-2iw=4iz\\ (3+2i)w-3w-2iw=4i(3+2i)\\ w(3+2i-3-2i)=-8+12i\\ w= \frac{-8+12i}{0}\\ \color{blue}w=indefinite$$

!

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Jan 28, 2024
edited by asinus  Jan 28, 2024
edited by asinus  Jan 29, 2024
#2
+259
0

We can rearrange the equation as follows:

zw - 3w - 2iw + 4iz = -8 + 12i

(z - 4i)(w - 2i) = -8 + 12i

z - 4i = -\frac{8 + 12i}{w - 2i}

z = -\frac{8 + 12i}{w - 2i} + 4i

This is a general formula for all solutions to the equation, where z and w are complex numbers.

To find specific solutions, we can substitute in any values for w. For example, if we let w=1, then we get:

z = -\frac{8 + 12i - 8i + 8}{1 + 2}

z = -\frac{4 + 4i}{3}

Therefore, one solution is z=âˆ’34+4iâ€‹. There are infinitely many other solutions, each of which can be found by substituting in a different value for w.

Jan 28, 2024
#3
+14826
0

zw - 3w - 2iw + 4iz = -8 + 12i

Where did you get this equation?

!

asinus  Jan 28, 2024