Complex Number Problem (1 Viewer)

[kev-kun]

Hi all, I just started the Ext.2 course yesterday and need help with this question:

<img src="http://latex.codecogs.com/gif.latex?z\in&space;\mathbb{C}$&space;such&space;that$&space;\frac{z}{z-1}&space;$&space;is&space;real.&space;Show&space;that&space;z&space;is&space;imaginary.$" title="z\in \mathbb{C}$ such that$ \frac{z}{z-1} $ is real. Show that z is imaginary.$" />

From Cambridge Textbook. Also on a side note, am I supposed to be able to solve this easily? =='

 

[SpiralFlex]

Hi all, I just started the Ext.2 course yesterday and need help with this question:

<img src="http://latex.codecogs.com/gif.latex?z\in&space;\mathbb{C}$&space;such&space;that$&space;\frac{z}{z-1}&space;$&space;is&space;real.&space;Show&space;that&space;z&space;is&space;imaginary.$" title="z\in \mathbb{C}$ such that$ \frac{z}{z-1} $ is real. Show that z is imaginary.$" />

From Cambridge Textbook. Also on a side note, am I supposed to be able to solve this easily? =='

What have you tried? If z is an element of the complex field, then z can be written in the form z=______. We also note that z/(z-1) is real. That means Im(z) = ?

The above is a common way, can you spot another method?

Edit: question is wrong see below. Methods still remain valid.

 

[QZP]

What have you tried? If z is an element of the complex field, then z can be written in the form z=______. We also note that z/(z-1) is real. That means Im(z) = ?

The above is a common way, can you spot another method?

Can you tell me where I'm going wrong:

If z is such that z/(z-1) is real, then
z/(z-1) = x, where x is real
z = x(z-1)
= xz - x

Therefore (x-1)z = x
Hence z = x/(x-1)

I still obtain z to be a real number D:

 

[seanieg89]

Hi all, I just started the Ext.2 course yesterday and need help with this question:

<img src="http://latex.codecogs.com/gif.latex?z\in&space;\mathbb{C}$&space;such&space;that$&space;\frac{z}{z-1}&space;$&space;is&space;real.&space;Show&space;that&space;z&space;is&space;imaginary.$" title="z\in \mathbb{C}$ such that$ \frac{z}{z-1} $ is real. Show that z is imaginary.$" />

From Cambridge Textbook. Also on a side note, am I supposed to be able to solve this easily? =='

Err, that's not true. Any real z that isn't 1 will give z/(z-1) real.

 

[seanieg89]

Can you tell me where I'm going wrong:

If z is such that z/(z-1) is real, then
z/(z-1) = x, where x is real
z = x(z-1)
= xz - x

Therefore (x-1)z = x
Hence z = x/(x-1)

I still obtain z to be a real number D:

Yeah you are right and the book is wrong, unless you typoed the question.

 

[QZP]

Yeah you are right and the book is wrong, unless you typoed the question.

Ah thanks for clarifying. Was worried for a few moments :S

 

[HeroicPandas]

Yeah you are right and the book is wrong, unless you typoed the question.

He did a typo, its meant to be

 

[seanieg89]

He did a typo, its meant to be

Cool, then his working pretty much does it. we have z/i must be real and hence z must be imaginary or zero. (book should have specified zero as a possibility).

 

[kev-kun]

Ooh crap yeah it was a typo. Supposed to be i instead of 1. Thanks for help... I kind of get it. So since z/z-i is real, you let it equal x and sub it into the form z=x+iy?