argument

[burgerandcheese]

Please tell me why it must be -π < θ <= π

I don't understand. if z = rcosθ + (rsinθ)i, z can only be 0 if cos θ=-1 and sin θ=1 or cos θ=-1 and sin θ=1

 

[Dr.Peterson]

That is part of the definition. Some interval has to be chosen in order to make "the argument" unique; they could have chosen 0 <= θ <= 2π instead.

If z = 0, what is r?

In that case, the argument is undefined because any value of θ would give the same z.

I don't understand your claim that "z can only be 0 if cos θ=-1 and sin θ=1 or cos θ=-1 and sin θ=1". Why do you say that? Those are impossible conditions, and wouldn't yield 0 anyway.

 

[Deleted member 4993]

View attachment 12520View attachment 12521

Please tell me why it must be -π < θ <= π

I don't understand. if z = rcosθ + (rsinθ)i, z can only be 0 if cos θ=-1 and sin θ=1 or cos θ=-1 and sin θ=1

From your attached figure:

y = r*sin(Θ) and x = r*cos(Θ)

You also know:

sin(Θ) = sin(Θ+2*π) and cos(Θ) = cos(Θ+2*π)

So after every 2π angle (cycle) - values keep repeating. Now think ......

 

[burgerandcheese]

Yes, sorry. I thought -π < θ <= π meant the argument is only in quadrants 1, 3 and 4. I thought pi radians was 90 degrees lol ?

I don't understand your claim that "z can only be 0 if cos θ=-1 and sin θ=1 or cos θ=-1 and sin θ=1". Why do you say that? Those are impossible conditions, and wouldn't yield 0 anyway.

Yes I forgot that cosθ and sinθ are multiplied by r