What is the argument of 1+i in the complex number 1/(1+i)?

In summary, the inverse of a complex number is another complex number that, when multiplied by the original complex number, results in a product of 1. To find the inverse of a complex number, you can use the formula z^-1 = (a - bi) / (a^2 + b^2), where a + bi is the original complex number and a - bi is the complex conjugate of the original number. The main difference between a complex number and its inverse is that the inverse has the opposite sign for the imaginary component. The inverse of a complex number can be a real number if the imaginary component of the original number is equal to 0. The geometric interpretation of the inverse of a complex number is that it is
  • #1
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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  • #2
mcastillo356 said:
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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  • #3
##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.
 
  • #4
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##.
 
  • #5
PeroK said:
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?
 
  • #6
mcastillo356 said:
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.
 
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  • #7
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
 
  • #8
There's nothing to add. The polar form of a complex number repeats every ##2\pi## just like sine and cosine.
 
  • #9
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.
 
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Thank you very much, Perok!, also thanks to you, Frodo!
 
  • #11
mcastillo356 said:
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.
 
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  • #12
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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1. What is the definition of the inverse of a complex number?

The inverse of a complex number is another complex number that, when multiplied by the original complex number, results in a product of 1. In other words, the inverse of a complex number is the number that "undoes" the original number when multiplied together.

2. How is the inverse of a complex number calculated?

The inverse of a complex number is calculated by taking the reciprocal of the complex number. This means that the real part of the original complex number becomes the denominator of the inverse, while the imaginary part becomes the numerator with a negative sign. For example, the inverse of the complex number 3+4i would be 1/(3+4i) = 3/25 - 4/25i.

3. Can every complex number have an inverse?

No, not every complex number has an inverse. A complex number only has an inverse if its magnitude (or absolute value) is not equal to 0. This means that complex numbers with a real and imaginary part of 0 (such as 0+0i) do not have an inverse.

4. How is the inverse of a complex number used in mathematics?

The inverse of a complex number is used in many areas of mathematics, including complex analysis, differential equations, and linear algebra. It is also used in physics and engineering to solve problems involving electrical circuits and waves.

5. What is the relationship between the inverse of a complex number and its conjugate?

The inverse of a complex number is related to its conjugate in that the product of a complex number and its conjugate is always a real number. This means that the inverse of a complex number can also be written as the conjugate divided by the magnitude squared. For example, the inverse of the complex number 3+4i can also be written as (3-4i)/(3^2+4^2) = (3-4i)/25.

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