Finding Modulus and Argument for a Complex Number

4 replies · 2K views · last reply
ChrisBaker8
Messages
23
Reaction score
0

Homework Statement



Find the modulus and the principal value of the argument for the complex number [tex]\sqrt{3}[/tex] - i

The Attempt at a Solution



I know the modulus is just 'square both, add, and square root of sum', so r = [tex]\sqrt{2}[/tex], but I don't know how to find the second part. I know vaguely that the argument = [tex]\theta[/tex], but I don't know where to go from here.

Do I need to convert the complex number into polar or euler form?
 
What two numbers did you 'square both, add, and square root of sum' to get sqrt(2)? The complex number a+bi can be drawn as the hypotenuse of a right triangle in the complex plane with a horizontal leg of length a and a vertical leg of length b. Haven't you seen this picture? The argument is the angle the hypotenuse makes with the x-axis. So you have tan(argument)=b/a. Remember trig? Principal value is a convention for choosing which of several possible angles might satisfy the tangent equation. Look it up, hopefully it will come with a nice picture.
 
okay, I think I get the argument now

for the modulus, I added [tex]\sqrt{3}[/tex] [tex]^{2}[/tex] and i[tex]^{2}[/tex] to get 3 - 1, then square rooted to get [tex]\sqrt{2}[/tex]

is that wrong?
 
ChrisBaker8 said:
okay, I think I get the argument now

for the modulus, I added [tex]\sqrt{3}[/tex] [tex]^{2}[/tex] and i[tex]^{2}[/tex] to get 3 - 1, then square rooted to get [tex]\sqrt{2}[/tex]

is that wrong?
Yes. You square the real part, sqrt(3), and the imaginary part, -1, add them, then take the square root. The imaginary part is the coefficient of i.