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

 

please help im so stuck 😭😭😭

 Jan 28, 2024
 #1
avatar+14826 
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\(zw-3w-2iw=4iz=-8+12i\)

 

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

 

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

 

laugh!

.
 Jan 28, 2024
edited by asinus  Jan 28, 2024
edited by asinus  Jan 29, 2024
 #2
avatar+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=−34+4i​. There are infinitely many other solutions, each of which can be found by substituting in a different value for w.

 Jan 28, 2024
 #3
avatar+14826 
0

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

Where did you get this equation?

laugh!

asinus  Jan 28, 2024

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