Find argument of complex number

Question

I want to find the argument of $$ \frac{6e^{-iT/2}}{(1+i)^2} $$ Attempt:

$\arg 6+\arg e^{-iT/2}-\arg((1+i)^2)$, where $\arg((1+i)^2)=\arg(2j)$. So $\arg 6+\arg e^{-iT/2}-\arg(2j)=\arctan(1/6)-T/2+\arctan(2)$.

Correct answer is $-T/2-2\arctan(1)=-T/2-\pi/2$.

What have I missed?

Qurultay · Accepted Answer · 2017-10-26 13:48:55Z

1

$\arg(1+i)^2=2\arg(1+i)=2 \frac{\pi}{4}=\frac{\pi}{2}$ and $\arg(6)=\arg(6+0i)=0 $ so, $$\arg\frac{6e^{-iT/2}}{(1+i)^2}=-T/2+\frac{3\pi}{4}. $$

answered Oct 26, 2017 at 13:48

Qurultay

5,2192 gold badges14 silver badges26 bronze badges

$\begingroup$ Hi! Why is $\arg(1+i)^2=2\arg(1+i)$? $\endgroup$
– JDoeDoe
Commented Oct 26, 2017 at 13:53
$\begingroup$ $\arg(1+i)^2=\arg 2i$ $\endgroup$
– user284001
Commented Oct 26, 2017 at 13:55
$\begingroup$ generally, if $z=\cos(\theta)+i\sin(\theta)$, then $z^n=\cos(n\theta)+i\sin(n\theta) $ i.e. $\arg(z^n)=n\arg(z) $ of corce with an appropriate change to confirm $0\leq \arg(z)\leq 2\pi$ if necessary. $\endgroup$
– Qurultay
Commented Oct 26, 2017 at 13:58

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Swapnil Rustagi · Accepted Answer · 2017-10-26 13:51:30Z

0

$\arg(z) = \arctan ( \frac{Im(z)}{Re(z)} )$

(We might need to add or subtract $\pi $ depending on which quadrant the complex number belongs to)

$\arg(6) = \arctan (0/6) = 0$

$\arg((1+i)^2) = \arg (2i) = \pi/2$

Also your final answer seems wrong: it should be $-T/2 - \pi/2$

answered Oct 26, 2017 at 13:44

Swapnil Rustagi

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2 Answers 2