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:

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

 

[blue_raver22]

To linearize the imaginary function, we need to rewrite it in the form of a+bi, where a is the real part and bi is the imaginary part. To do this, we can use the fact that i^2 = -1.

First, let's simplify the given function:

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

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

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

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

Next, we can multiply the numerator and denominator by the complex conjugate of the denominator, which is -4+3i:

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

= (-8-6i+4i-3i^2)/(-16-12i+12i-9i^2)

= (-8-2i)/(-16+9)

= (-8-2i)/(-7)

Now, we can rewrite this in the form of a+bi:

z = (-8-2i)/(-7)

= (-8/-7) + (-2/-7)i

= (8/7) + (2/7)i

Therefore, the real part (Re(z)) is 8/7 and the imaginary part (Im(z)) is 2/7. This is different from the answer given by Wolfram because we have simplified the function differently. However, both answers are equivalent and correct.

To check if the two equations you have equated are equal, we can plug in the values for z that we found above:

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

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

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

= (2-i)/(-1+4.8i)

= (2-i)/(-1+4.8i)

= (2-i)/(4.8i-1)

= (2-i)/(4.8i-1)

= (2-i)/(4.8i-1)

= (2-i)/(4.8i-1)

= (2-i)/(4.8