How Can I Linearize the Complex Function z = (2+i)/(i(-3+4i))?

[craig16]

so i have the function z=(2+i)/(i(-3+4i)) and i need to linearize it to find the Im(z) and Re(z)

I get down to z= (-6 +8i -3i -4 )/ (9i +12 +12 -16i) which i then simplify down to

z= (5i -10)/(-5i+24)

However when solve it i get a different answer from wolfram (from when i plugged z=(2+i)/(i(-3+4i))).

And when i try to equate

z= (-6 +8i -3i -4 )/ (9i +12 +12 -16i) and z= (5i -10)/(-5i+24)

wolfram tells me they are not equal. How do i do this question?

 

[I like Serena]

so i have the function z=(2+i)/(i(-3+4i)) and i need to linearize it to find the Im(z) and Re(z)

I get down to z= (-6 +8i -3i -4 )/ (9i +12 +12 -16i)

How did you get this?

 

[craig16]

Separate the top of the original function, cancels out an i.
Then make a common denominator and add the top of both functions.

z= 2/i(-3+4i) + i/i(-3+4i)

z= 2/(-3i-4) + 1/(-3+4i)

z= 2(-3+4i) / (-3i-4)(-3+4i) + 1(-3i-4)/(-3i-4)(-3-4i)

etc..

 

[I like Serena]

so i have the function z=(2+i)/(i(-3+4i)) and i need to linearize it to find the Im(z) and Re(z)

I get down to z= (-6 +8i -3i -4 )/ (9i +12 +12 -16i) which i then simplify down to

z= (5i -10)/(-5i+24)

Well, in the step above, the denominator should not be (-5i+24) but (-7i+24).

Anyway, there is an easier way to do this using the equality (a+b)(a-b)=a²-b²:

z=\frac {2+i} {i(-3+4i)}=\frac {2+i} {-4-3i}=\frac {(2+i)(-4+3i)} {(-4-3i)(-4+3i)}=\frac {(2+i)(-4+3i)} {(-4)^2-(3i)^2}=\frac {(2+i)(-4+3i)} {16+9}=...