# Inverse of a complex number

• mcastillo356
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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!

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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.

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##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.

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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##.

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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?

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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
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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

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

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

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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.

mcastillo356
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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.

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