Mind Blank with this Q PLs Help Appreciated Thanks (1 Viewer)

[Modern4DaBois]

 

[jimmysmith560]

You may wish to have a look at the following thread where this question was asked in order to gain insight into this question:

complex number question

how do i solve this : Suppose that the complex number z has modulus one, and that 0 < Arg z < π/2 . Prove that 2 Arg(z + 1) = Arg z.

boredofstudies.org

I hope this helps!

 

[Modern4DaBois]

Ah ok. Conceptually that explanation makes sense. But how would you go about the working out? Are explanations enough, or do you have to set it up like a proper proof question?

 

[jimmysmith560]

Perhaps you need to go beyond explanations, as seen in the following working:

Step 1:

Given the complex number z has modulus one, and 0 < Arg z < π/2.

So, we may take complex number z as

z = cost + i sint.

Then |z| = 1 and

Argz = tan^-1(sint/cost) [since, 0 < Argz < π/2]
= tan^-1(tant)
= t [since, 0 <Argz <π/2 (given)]

Step 2:

z = cost +i sint,

Therefore,

z +1 = 1 + cost + i sint
= 2cos²(t/2) + i 2sin(t/2) cos(t/2)

[Since, 1+cost= 2cos²(t/2), sint=2sin(t/2)cos(t/2)]

Arg(z+1) = tan^-1[{ 2sin(t/2)cos(t/2)}/{2cos2(t/2)}] [since, 0 < Argz < π/2 (given)]
= tan^-1{tan(t/2)}
= t/2 [since, 0 < Argz < π/2 (given)]

or, 2Arg(z+1) = t = Arg(z) (Proved)

 

[Modern4DaBois]

Thank you so much Jimmy!!! Really appreciate the effort you put in

 

[yanujw]

Alternatively, by a geometrical method;

This diagram illustrates the situation. Then you must prove the two angles in red are equal because the bottom angle in red is arg(z+1), and the two angles combined are arg(z), making Arg(z) = 2arg(z+1)

The lower red angle is equal to the green angle by alternate angles. Then, since |z| = 1, the triangle bounded by the points 0, z and z+1 is isosceles. The green angle is equal to the upper red angle.

 

[Modern4DaBois]

Oh ok that makes sense as well. Thank you so much for your help m8! Really appreciate it