Why Does My Calculation of h(z) = Re(z) / Im(z) Yield -12i Instead of 12?

[pistolpete333]

Hi, so for a homework problem I have to evaluate these complex functions. The one I am having trouble on is:

evaluate h(z) = Re(z) / Im(z) where z = (5-2i) / (2 - i)

The answer is in the back of the book, which says that the solution is 12, however I keep getting -12i for my answer. I just need someone to tell me where am going wrong.

I first multiply z by (2 + i) / (2 + i) in order to move the imaginary number to the top, which leaves me with (12 + i) / 5, I then take the real part, 12 / 5 , and put it over the imaginary part, i / 5, and simplify this to 12 / i , where I multiply this by i / i , which leaves me with -12i

The answer is supposed to be 12, so if anyone could shed some light on this that would be great. Thanks everyone

 

[LCKurtz]

The imaginary part of 12/5 + (1/5)i is 1/5.

 

[Ray Vickson]

Hi, so for a homework problem I have to evaluate these complex functions. The one I am having trouble on is:

evaluate h(z) = Re(z) / Im(z) where z = (5-2i) / (2 - i)

The answer is in the back of the book, which says that the solution is 12, however I keep getting -12i for my answer. I just need someone to tell me where am going wrong.

I first multiply z by (2 + i) / (2 + i) in order to move the imaginary number to the top, which leaves me with (12 + i) / 5, I then take the real part, 12 / 5 , and put it over the imaginary part, i / 5, and simplify this to 12 / i , where I multiply this by i / i , which leaves me with -12i

The answer is supposed to be 12, so if anyone could shed some light on this that would be great. Thanks everyone

Re(z) and Im(z) are both REAL, so their ratio must be real as well.

 

[pistolpete333]

Wow that was pretty obvious. I knew I was messing something up, thanks guys