Question

Compute the argument of each complex number:
1 + i
sqrt(3) + i
-2i
5 - 5i
7sqrt(3) + 7i
-3 + 4i
-5

Answer 1

@baller398 @radar @Kentekai

Answer 2

....okay. Can you tag someone else who you think will help me?

Answer 3

@dan815 @Nnesha

Answer 4

Just remember that \(\arg(~a+bi~) = \sqrt{a^2+b^2}\)

Answer 5

I don't get it... example???

Answer 6

For example, first complex number; \(1+i\)

So \(\arg(~1+i~)=\arg(~1+1i~)=\sqrt{1^2+1^2}= \sqrt{2}\)

Answer 7

arg( a+bi ) stands for argument of a+bi, just so you know

Answer 8

But I got \[\sqrt2\] as the modulus...

Answer 9

is it okay to have to modulus and argument be the same value?

Answer 10

oh arg is angle...

Then you have \(\arg(~a+bi~) = \tan^{-1}\left(\dfrac{b}{a}\right)\)