What is the Principal Argument of -i in Polar Form?

[jernobyl]

Homework Statement

Express -i in polar form, using the principal value of the argument.

Homework Equations

modulus = \sqrt{a^2 + b^2}

\theta = arg(0 - i)

The Attempt at a Solution

Well, the complex number is 0 -i. a = 0, b = -1 so:

r = \sqrt{0^2 + (-1)^2} which comes out to be 1.

But for the argument, \theta comes out to be:

\tan\theta = \frac{-1}{0}

Ummm...where do we go from here?! Also, err, what IS the principal argument of the argument? I mean, it seems to be that the value of \theta changes depending not where on the CAST diagram it is, but on where in the Argand diagram the complex number turns out to be...but, err, not always. Like, if it lies in the fourth quadrant, you don't do 360 - \theta...

 

[cristo]

The argument of a complex number x+iy is only tan(y/x) if neither x nor y is zero.

You can find the argument of -i by simply considering the argand diagram. Where does -i lie on the argand diagram? When you plot this, it should be obvious what the argument is. Note that the principal argument, is a value between -pi and pi. (or -180 and 180 degees)

 

[jernobyl]

Thanks for this. The exercises I have been doing on complex numbers have all asked for the principal argument, and some of the answers are not between \pi and -\pi, such as 239.04 degrees.

Okay, err, I've plotted -i on the argand diagram...-i is at "-1"...I'm really not getting it here...

 

[cristo]

Thanks for this. The exercises I have been doing on complex numbers have all asked for the principal argument, and some of the answers are not between \pi and -\pi, such as 239.04 degrees.

you may have a different definition of the principal argument then; although i thought it was always between pi and -pi

Okay, err, I've plotted -i on the argand diagram...-i is at "-1"...I'm really not getting it here...

It's at -1 on the imaginary axis. Now, what is the angle between the positive real axis and the negative imaginary axis? This will give you the argument.

 

[D H]

Some alternatives are

• Use Euler's formula, e^{i\theta} = \cos \theta + i\sin \theta. You need to find the value \theta such that \cos \theta = 0 and \sin\theta = -1.
• Use the cotangent instead of the tangent, paying attention to the sign of the argument.
• Use the two argument form of the inverse tangent, http://en.wikipedia.org/wiki/Arctangent#Two_argument_variant_of_arctangent"

 

[rsnd]

its 1 e^ {-I * Pi/2} ?

 

[cristo]

its 1 e^ {-I * Pi/2} ?

... please don't simply give out answers to homework questions.

 

[jernobyl]

Ah, thanks for that. I'm confused but I kinda understand the answer...

 

[HallsofIvy]

Since you say some, at least of the answers are in degrees, between 0 and 360 degrees, start at the positive real axis and measure the angle counterclockwise to the negative y-axis. You should see that the angle is a multiple of 90 degrees.

 

[jernobyl]

Yeah, thanks. I'm just not sure, like...one of the answers to one of the questions was -7.15 degrees...and how can that be, if you're meant to measure it counterclockwise from the positive real axis? Sigh.