Moved - Arg of complex no.

[Anthonyk2013]

I edited my post..Do you understand it now or shall I explain it to you??

I understand it now thanks, only a few days at complex numbers.

Also have this

W=1+2i

Arg=(w-3i)=((1+2i)-3i)=

not sure if I'm have to subtract 2i-3i or multiply by each other?

 

[evinda]

\(\displaystyle ((1+2i)-3i)=(1-i) \)
\(\displaystyle tan\theta=\frac{y}{x}=\frac{-1}{1}=-1 \)
\(\displaystyle arg(W-3i)=\theta=tan^{-1}(-1)=-\frac{\pi}{4} \)

 

[Anthonyk2013]

\(\displaystyle ((1+2i)-3i)=(1-i) \)
\(\displaystyle tan\theta=\frac{y}{x}=\frac{-1}{1}=-1 \)
\(\displaystyle arg(W-3i)=\theta=tan^{-1}(-1)=-\frac{\pi}{4} \)

Im afraid you have lost me now, have not used tan yet. You know any good YouTube clips or web site to explain.

 

[evinda]

With arg don't you mean the angle??How did you get taught it??

 

[Anthonyk2013]

With arg don't you mean the angle??How did you get taught it??

Starting night class in two weeks. Trying to get a head start. Find maths very hard to take in so want to have some understanding before I start.

Taught from YouTube mainly over the last few days. I know I might be mad but just lost my job of 20 years so taking night classes. Need every advantage I can get.

Thanks.

 

[Anthonyk2013]

\(\displaystyle ((1+2i)-3i)=(1-i) \)
\(\displaystyle tan\theta=\frac{y}{x}=\frac{-1}{1}=-1 \)
\(\displaystyle arg(W-3i)=\theta=tan^{-1}(-1)=-\frac{\pi}{4} \)

Tan o = x
Y Do you treat (1-i) as(x+y) and solve from there?

 

[evinda]

The general form of a complex number is z=x+yi.
In your case,z=1+(-1)i
So x=1 and y=-1..

 

[Anthonyk2013]

The general form of a complex number is z=x+yi.
In your case,z=1+(-1)i
So x=1 and y=-1..

Thanks it's beginning to sink in.