# Homework Help: Complex number

1. Aug 17, 2010

### thereddevils

1. The problem statement, all variables and given/known data

Given that arg(z/(1-i))=pi/2, find the argument of z.

Sketch , in the Argand diagram , the set of points representing z and the point representing the complex number w=-4+3i. Hence deduce the least value of |z+4-3i|

2. Relevant equations

3. The attempt at a solution

Let z=a+bi

Rationalising z/(1-i)=(a-b)/2+(a+b)/2 i

negative pi/2 suggest that the point is on the negative y-axis.

tan pi/2 = [(a-b)/2]/[(a+b)/2]

hence a=b

And since (a+b)/2 is negative, a<0 , b<0

so z=a+ai or b+bi

(a,a) is in the third quadrant, and the argument of z is -3pi/4

I have no problem in sketching. The problem i have is with the last part.

Any pointers?

2. Aug 17, 2010

### Mentallic

So which is it, $\pi/2$ or $-\pi/2$?

You can solve the first faster and easier by using $$arg\left(\frac{a}{b}\right)=arg(a)-arg(b)$$

The least value of $$|z+4-3i|$$ is saying what point z where $$arg(z)=\frac{-3\pi}{4}$$ thus z=a(1+i), a<0 is closest to the point (-4+3i)? If you think about this in cartesian coordinates, z lies on the line y=x where x<0. Now all you have to do is find out the closest distance between the point (-4,3) and the line y=x.
The problem could even be simplified using logic since we know the line connecting the shortest distance between z and the point is at right angles so the gradient of the line is -1.

3. Aug 17, 2010

### thereddevils

thanks Mentallic.

4. Aug 17, 2010

### Mentallic

It sure does

If you want to prove it, let $$x=r cis\theta$$ and $$y=Rcis\phi$$ and then rationalize x/y and use some properties of trigonometry to simplify it down.

5. Aug 18, 2010

### HallsofIvy

It is precisely a "property of logarithms" because the argument of a complex number is a property of exponentials.

Write complex number a, with magnitude $r_1$ and argument $\theta_1$, in the form $r_1e^{i\theta_1}$ and the complex number b, with magnitude $r_2$ and argument $\theta_2$, in the form $r_2e^{i\theta_2}$.

Then
$$\frac{a}{b}= \frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}= \frac{r_1}{r_2}\frac{e^{i\theta_1}}{e^{i\theta_2}}= \frac{r_1}{r_2}e^{i(\theta_1- \theta_2)}}$$

That is, the magnitude of a/b is $r_1/r_2$ and its argument is $\theta_1- \theta_2$.

6. Aug 18, 2010

### Dickfore

And another student discovers the miracles of Euler's formula.

7. Aug 18, 2010

### thereddevils

Thanks Hallsofivy, and wow how long have you been on this forum?You have made 27000 posts!

8. Aug 22, 2010

### thereddevils

Is the answer for the least value sqrt(53/2) ?

9. Aug 22, 2010

### Mentallic

No not quite. Show us what you did.

10. Aug 22, 2010

### thereddevils

I drew a diagram, with right angled triangle WOZ. Using the perpendicular gradients,

(a/a)((3-a)/(-4-a))=-1

a=-1/2

Then i proceed to find the distance between (-4,3) and (-1/2,-1/2) which gives that answer.

11. Aug 22, 2010

### Mentallic

Ok so you set up your question properly, you just used the distance formula incorrectly.

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Where $$(x_1,y_1)=(-1/2,-1/2)$$ and $$(x_2,y_2)=(-4,3)$$

I can think of some other ways to solve this problem which might be of interest to you.

So we have $$d=\sqrt{(x-4)^2+(y+3)^2}$$ to denote the distance from the point(4,-3) to a circle's circumference with that centre and radius d. We also have the equation y=x so if we plug y into the distance equation and simplify we get $$d=\sqrt{2x^2-2x+25}$$ but we want to find the minimum distance so we want the minimum value of the quadratic under the sq.root. Just use some calculus and you'll find x=1/2, which can be substituted back into the distance equation to give you the answer.

12. Aug 22, 2010

### thereddevils

thanks Mentallic, it can also be solved using trigonometry and vectors.

13. Aug 22, 2010

### Mentallic

Yep, and if you go back to precalculus class they would be using a specific distance formula for the distance between a point and a line

By the way, I have no idea why this was moved to the precalc forum.

14. Aug 22, 2010

### HallsofIvy

I started last night and, boy, are my fingers tired!:rofl:

15. Aug 22, 2010

### Mentallic

Halls, an average of 10 posts per day for the passed 7 years... umm... just one question. Assuming you have to take a break once a year or so, for christmas maybe how many posts have you made on your big days? I'm suspecting you spammed the general forum like hell :tongue: