How Do I Solve Complex Equations with Different Lambda Values?

[Banker]

Homework Statement

Question 3.b. - http://imgur.com/ztLiRvx

Homework Equations

For the sake of simplicity, let's assume that lambda = x.

The Attempt at a Solution

I tried equating the real an imaginary parts of arctan(1/4).
Real: x/2 + 3 = 4. This gives x = 2.
Imaginary: x/2 - 3 = 1. This gives x = 8.
I'm getting two different values of x(lambda).

 

[SammyS]

Homework Statement

Question 3.b. - http://imgur.com/ztLiRvx

Homework Equations

For the sake of simplicity, let's assume that lambda = x.

The Attempt at a Solution

I tried equating the real an imaginary parts of arctan(1/4).
Real: x/2 + 3 = 4. This gives x = 2.
Imaginary: x/2 - 3 = 1. This gives x = 8.
I'm getting two different values of x(lambda).

$\displaystyle arg(z) = tan^{-1}\left(\frac{Im(z)}{Re(z)}\right) \$

so you only know the ratio of the real to imaginary parts. It's not that the real part is 4 and the imaginary part is 1 .

 

[Mark44]

Homework Statement

Question 3.b. - http://imgur.com/ztLiRvx

Homework Equations

For the sake of simplicity, let's assume that lambda = x.

The Attempt at a Solution

I tried equating the real an imaginary parts of arctan(1/4).

?
I don't know what you're doing. arctan(1/4) means the angle (number) whose tangent is 1/4.

Real: x/2 + 3 = 4. This gives x = 2.
Imaginary: x/2 - 3 = 1. This gives x = 8.

No.
If z = x I iy, then arg(z) = arctan(y/x).

Use the result from part a of this problem and the above to find $\lambda$ -- there is a single value.

I'm getting two different values of x(lambda).

 

[Banker]

@Mark44 In my original attempt, I tried to reverse the process you mentioned, but as Sammy pointed out, it only tells the ratio.
So I then divided the equations for the imaginary and real parts in part 'a' and got 10 as my answer. So that will give 68 as the answer to the second part of 3.b, right?

 

[SammyS]

@Mark44 In my original attempt, I tried to reverse the process you mentioned, but as Sammy pointed out, it only tells the ratio.
So I then divided the equations for the imaginary and real parts in part 'a' and got 10 as my answer. So that will give 68 as the answer to the second part of 3.b, right?

Yes, those are correct, although we generally like to see details of how you worked it out. That makes it easier for us to check your answer.

By the way, if you click on the big " Σ " in the blue menu bar at top of the message window, you will have easy access to many characters, including λ . Otherwise, it would be wise to use almost any other letter to represent λ, other than x, y, or z . Upper case L would have been a good choice.

 

[Banker]

@SammyS
Thanks man, here is my working - http://imgur.com/fkk25DE

 

[SammyS]

@SammyS
Thanks man, here is my working - http://imgur.com/fkk25DE

Looks good.