Finding arg(iz) when z = 2 - i

[patm95]

Homework Statement

I need help on a little review please. z=2-i What is arg(iz)

Homework Equations

Well iz= 1+2i

The Attempt at a Solution

I think this should end up being arg(2i/1) But this doesn't seem to make sense because I am wanting to find an angle here right? I am somewhat confused. This is supposed to be an inverse tangent type problem, correct?

 

[patm95]

I am thinking now that this should be something like 3pi/8 but I can't explain my logic...

 

[Mentallic]

It's simple trigonometry if you take a look at a complex number on the argand diagram. The argument of the complex number is the angle between the vector and the x-axis. The vector is just a line between the origin and the complex number.

So, $$arg(a+ib)=tan^{-1}(b/a)$$

 

[patm95]

Ok. Doesn't that mean that it would be tan-1(2) which would be 1.107?

 

[HallsofIvy]

No, it doesn't! The arg of a+ bi is the angle the line from 0 to a+bi, thought of as points in the complex plane (a+ bi-> (a, b)) makes with the positive real axis. Specifically it is, as $tan^{-1}(b/a)$ as Mentallic says.

But your number is z= 2- i. a= 2 and b= -1, NOT a= 1 and b= 2 as you seem to think.

 

[patm95]

Yeah but the question is what is What is arg(iz) so doesn't that mean that I mult i times 2-i to give me 1+2i?

I think what you just said would be for arg(z). correct?

 

[HallsofIvy]

Sorry, my fault, I didn't read the problem carefully enough. If z= 2- i then iz= 2i+ 1 so, yes, the arg of 1+ 2i is, in fact, $tan^{-1}(2)$.

 

[patm95]

Ok. Thanks!