Arg vs arg: Let z = (1 - 7i)/(3 + 4i). Express z in the form a + bi. Find Arg z^2.

[0.5KV^2]

Hi All,

Sorry if this is on the wrong thread, I am not sure where complex numbers go and I am not an American student!

I am stuck on part ii of this question. I got -1 - i as my answer for part i. I am confused whether the Arg (z) should be -3pi/2 or pi/2. I understand that if you squared the z value you get 2i, which then gives the Arg value as pi/2. But if I took the 2 out and made it 2Arg(z) then I would get -3pi/2. I understand that capital 'A' in arg indicates the principle argument, but both these values lie between the boundaries for this?

Thanks

 

[blamocur]

pi/2 and -3*pi/2 are equivalent since the difference between them is 2*pi. What is the definition of "principle argument" ?

 

[Harry_the_cat]

pi/2 and -3*pi/2 are equivalent since the difference between them is 2*pi. What is the definition of "principle argument" ?

"coterminal" not "equivalent"

 

[Dr.Peterson]

I understand that capital 'A' in arg indicates the principle argument, but both these values lie between the boundaries for this?

The interval for the principal value of the argument is defined in a couple different ways. But I don't know of any that would include both pi/2 and -3*pi/2. So we do need to see how you have been taught.

 

[blamocur]

"coterminal" not "equivalent"

Isn't coterminality an equivalence relation

 

[pka]

Hi All,

Sorry if this is on the wrong thread, I am not sure where complex numbers go and I am not an American student!

I am stuck on part ii of this question. I got -1 - i as my answer for part i. I am confused whether the Arg (z) should be -3pi/2 or pi/2. I understand that if you squared the z value you get 2i, which then gives the Arg value as pi/2. But if I took the 2 out and made it 2Arg(z) then I would get -3pi/2. I understand that capital 'A' in arg indicates the principle argument, but both these values lie between the boundaries for this?

View attachment 36544

If [imath]~z~[/imath] is a complex number then [imath]\large\left( {\dfrac{1}{z} = \dfrac{{\overline{~ z~} }}{{|z{|^2}}}} \right)[/imath]. Using that we get [imath]\dfrac{{1 - 7i}}{{3 + 4i}}\left( {\dfrac{{3 - 4i}}{{3 - 4i}}} \right) = \dfrac{{3 - 21i - 4i - 28}}{{25}} = - 1 - i[/imath]
[imath]\large{ \arg ( - 1 - i)=-\pi+\frac{\pi}{4}}[/imath]

Now for your question: Suppose that [imath]z=a+b{\bf{i}}~\&~a\cdot b\ne 0[/imath]. You must be able to tell the quadrant [imath]z[/imath]is in.
Next find [imath]\tau = \arctan \left( {\left| {\dfrac{b}{a}} \right|} \right)[/imath], and then [imath] \arg (z) = \theta = \left\{ \begin{gathered} \tau ,\quad (x,y) \in I \\ \pi - \tau ,\quad (x,y) \in II \\ - \pi + \tau ,\quad (x,y) \in III \\ - \tau ,\quad (x,y) \in IV \\ \end{gathered} \right.[/imath]

So what is the argument of [imath]\left( {\dfrac{{1 - 7i}}{{3+ 4i}}} \right)^2[/imath]