# Argument of a random complex no. lying on given line segment.

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1. Dec 16, 2015

### Ricky_15

1. The problem statement, all variables and given/known data
In the argand plane z lies on the line segment joining # z_1 = -3 + 5i # and # z_2 = -5 - 3i # . Find the most suitable answer from the following options .

A) -3∏/4

B) ∏/4

C) 5∏/6

D) ∏/6

2. MY ATTEMPT AT THE SOLUTION

We get two points ( -3 , 5 ) & ( -5 , -3 ) => The line segment must intersect x - axis and lie in the 2nd and 3rd quadrant .
=> #z# lies on x-axis or 2nd or 3rd qudrant .

But , since there is no option ∏ , so z must be lying in 2nd or 3rd quadrant.

=> -3∏/4 or 5∏/6 should be the solution .

I cant proceed further from here so as to differentiate between the above two choices.

The answer in the book is : 5∏/6 .

Last edited: Dec 16, 2015
2. Dec 16, 2015

### haruspex

The set of choices is only defined as far as directions from the origin. So I would think the approach is to find the arguments of z1 and z2.

Last edited: Dec 17, 2015
3. Dec 17, 2015

### LCKurtz

How is your line situated with respect to the line $y = x$, which is where a $z$ with argument $-\frac{3\pi}{4}$ would be? Draw a graph.

4. Dec 18, 2015

### jk22

You could also compute the argument : z=end-start and arg z or atan dy/dx

5. Dec 18, 2015

### BvU

Hello Rick,

If you pick random points on the line between those two points on your drawing, the average postion of those points should end up where, do you think ? You calculated the intersection with the negative x-axis, but you should have calculated the midpoint of the line segment. The angle that is closest to the argument of that point is your best answer.

Oh, and: you did draw a graph, I hope ?

6. Dec 18, 2015

### SammyS

Staff Emeritus
Is that a complete statement of the problem - word for word?

If not, please give a complete statement of the problem - word for word. Also, if a portion of your question is in the thread title, please include it in the body of the thread as well.

7. Dec 18, 2015

### LCKurtz

Anyone want to give odds on the OP ever returning to this thread?

8. Dec 18, 2015

### SammyS

Staff Emeritus
$\displaystyle \ \left(\frac{1}{e}\right)^\pi \$

9. Dec 18, 2015

### haruspex

That's completely irrational. Probably.

10. Dec 18, 2015

### Ricky_15

I am extremely sorry . I was in a hurry . Thats not the complete question . The question asked for " find the suitable solution for arg(z) " .

11. Dec 18, 2015

### haruspex

Have you tried my suggestion in post #2?

12. Dec 18, 2015

### Ricky_15

Yup that solved it , but for that I needed a calculator to find tan inverse 3/5 & 5/3 .