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Solve for $z$ in the following equation: $2-3iz = 3 + 2z$.

 Mar 6, 2024
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z = a  + bi

 

2  - 3i(a + bi)    = 3 + 2(a + bi)

 

2 - 3ai - 3bi^2  =  3 + 2a + 2bi

 

2 - 3ai + 3b  =  3 + 2a + 2bi

 

-3ai =  2bi

-3a = 2b

a = (-2/3)b

 

2 - 3(-2/3)b i  + 3b  = 3 + 2(-2/3)b + 2bi

2 + 3b  = 3 - (4/3)b

(13/3)b =  1

b = 3/13

 

a = (-2/3)(3/13)  = -2/13

 

z = -2/13  + (3/13)i

 

cool cool cool

 Mar 6, 2024

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