Inverse of a complex number

[mcastillo356]

Homework Statement

How can we divide $\dfrac{1}{1+i}$ and which is the argument?

Relevant Equations

$\dfrac{1}{w}=\dfrac{\overline{w}}{w\cdot{\overline{w}}}$

$\dfrac{1}{1+i}=\dfrac{1-i}{1-(-1)}=\dfrac{1}{2}-\dfrac{1}{2}i$. But the argument of $\dfrac{1}{1+i}$? I mean, why is that of $1+i$? Why $1+i\Rightarrow tg(\alpha)=\dfrac{1}{1}=1$?
Greetings!

 

[PeroK]

Homework Statement:: How can we divide $\dfrac{1}{1+i}$ and which is the argument?
Relevant Equations:: $\dfrac{1}{w}=\dfrac{\overline{w}}{w\cdot{\overline{w}}}$

$\dfrac{1}{1+i}=\dfrac{1-i}{1-(-1)}=\dfrac{1}{2}-\dfrac{1}{2}i$. But the argument of $\dfrac{1}{1+i}$? I mean, why is that of $1+i$? Why $1+i\Rightarrow tg(\alpha)=\dfrac{1}{1}=1$?
Greetings!

It's not clear what you don't understand here?

One idea is to express $1 + i$ in polar form and then you can see immediately what $\frac 1 {1 + i}$ must be in polar form.

 

[mcastillo356]

$1+i=\sqrt{2}_{\pi/4}$, isn't it?.
$\dfrac{1}{1+i}=r_{\alpha}$
$r=\sqrt{\dfrac{1}{2}}=\dfrac{1}{\sqrt{2}}$
$\alpha=\arctan{\dfrac{-1/2}{1/2}}=-\dfrac{\pi}{4}$
I'm I right?. $1+i$ and $\dfrac{1}{1+i}$ have different modulus and arguments. And I thought for a moment $1+i$ and $\dfrac{1}{1+i}$ had the same argument, to my surprise; that's why I've posted. It's been my mistake. I apologize.

 

[PeroK]

Note that if $z = re^{i\theta}$, then $\frac 1 z = \frac 1 r e^{-i\theta}$.

Subject to possibly adjusting $-\theta$ to lie between $0$ and $2\pi$.

 

[mcastillo356]

Note that if z=reiθ, then 1z=1re−iθ.

Subject to possibly adjusting −θ to lie between 0 and 2π.

Ok, but, what is the reason to make it lie in that range? To make a biyection? To make unique the argument?

 

[PeroK]

Ok, but, what is the reason to make it lie in that range? To make a biyection? To make unique the argument?

You should have learned about this. You need a convention for the standard representation of a complex number. It's either $\theta \in [0, 2\pi)$ or $\theta \in (-\pi, \pi]$. And, yes, a convention is needed to make the argument unique.

 

[mcastillo356]

Fine! I was only wondering if you wanted to add something about it: I'm infront of a lack of a definite textbook. Functions like sin, cos, and tan, are not biyective, but if we restrain the domain ($\mathbb{R}$) to a $2\pi$ amplitude's interval, then they are biyective and then we can consider the trigonometric inverse functions arcsin, arccos, arctan. That's what I've understood quite well

 

[PeroK]

There's nothing to add. The polar form of a complex number repeats every $2\pi$ just like sine and cosine.

 

[Frodo]

The book is multiplying both top and bottom by (1 - i) to get (1 - i) / (1 + i)(1 - i). Multiplying out the bottom line gives 2.

The answer is therefore (1 - i)/2.

 

[mcastillo356]

Thank you very much, Perok!, also thanks to you, Frodo!

 

[FactChecker]

Ok, but, what is the reason to make it lie in that range? To make a biyection? To make unique the argument?

It is just convenient to use a standard representation of the angle in $[0,2\pi)$ or [0, 360 deg). It is usually better to say that an angle is 38 degrees than to say that it is -4282 (= 38 -12*360). But there may be times when other ranges are more convenient. In fact, there are times when a person might want to track a continuous angle and not restrict the range at all. Those are unusual.

 

[Frodo]

In general, whenever you see (a + b) or (a - b) in a maths problem it's always worth checking to see if you can multiply it by the other to get a^2 + b^2.

The fact that (a + b) * (a - b) = a^2 - b^2 is surprisingly often useful.