Complex numbers Definition and 724 Threads

I understand that the "spiral" converges to 1+i-1/2-i/3!+1/4!+i/5!-1/6!-i/7!... . It splits into two: one for Re, 1-1/2+1/4!-1/6!..., and the other for Im, 1-1/3!+1/5!-1/7!... . Any hints on how to compute them?

For context, I have a 2x2 system of linear first order differential equations $$\vec{x}'=A\vec{x}$$ where $$A=\begin{bmatrix} 0 & 1\\c&-2\end{bmatrix}$$ and the characteristic polynomial is $$\lambda^2+2\lambda-c=0$$ The two eigenvalues are ##-1\pm\sqrt{1+c}##. Suppose ##c<-1##. Then...

I know that the definition of multiplication for integers is just repeated addition. For example, 5 times 3 means 5 + 5 + 5, but what about if we want to extend this definition to real or complex numbers ? Like for example, what does pi times e mean ? How are we supposed to add pi to itself e...

In my lines i have, ##(4-2i)^2 = (4-2i)(4-2i)## ##r^2 = 4^2 + (-2)^2 = 20## ##r \cos θ = 4## and ##r\sin θ = -2## ##\tan θ =-\dfrac{1}{2}## ##⇒θ = 5.82## radians. Therefore, ##|(4-2i)^2| = \sqrt{20} ×\sqrt{20} = 20## Argument = ##5.82 + 5.82 = 11.64##. also ##|2|## = ##2## and argument =...

To solve part (a), we write ##e^{inx}e^{-imx}=e^{ix(n-m)}##. If ##m=n## then this expression is 1, and so the integral of 1 from 0 to ##2\pi## is ##2\pi##. If ##m\neq n## then we use Euler's formula and integrate. The result is zero. My question is how do we solve part (b) using part (a)? I...

There is that term ##5^{1/3}## but that is exactly what we're trying to find.

Firstly, the exercise itself is not difficult: On one hand, $$|(a + ib)(c + id)|^2 = |a + ib|^2|c + id|^2 = (a^2 + b^2) (c^2 + d^2) = MN.$$ On the other hand, ##(a + ib)(c + id) = p+ iq## for some integers p and q, and so $$|(a + ib)(c + id)|^2 = |p + iq|^2 = p^2 + q^2.$$ Thus, ##MN = p^2 +...

In circuit analysis, everything seems to work out when you set i*cos = sin. But thats not a legitimate equation, so why does that work? Is there a proof that this is a real equation?

Hi! I am ok with understanding Euler's formula and how its proven. It is basic mathematic operations that are made possible by the characteristics of i, cos, sin, and exp. What still makes me uncomfortable is the jump we make at the very beginning or end of calculations, basically Acosx <==>...

I was thinking of investigating field theory because i like reading about quantum interpretations. What role does complex numbers have in physics? I have a hard time seeing why properties of a point in that field are not just multi dimensional properties on some parameter space? Why start...

If I have 2 measurements ##x = (3.0 ± 0.1), y = (-2.0 ± 0.1)## and want to calculate how the error propagates when calculating a function from those values this formula should be used: ##f(x, y) = f(x, y) ± \sqrt {(\frac{\partial f}{\partial x}*\Delta x)^2+(\frac{\partial f}{\partial y}*\Delta...

Hello, I have a question about the FFT and would like to share my thoughts with you. The background is a problem 30.2-6 from the legendary algorithms book by Cormen. It says that instead of doing an n element FFT over the field of complex numbers, we can use ##\mathbb{Z}_m##, where ##m = 2^{t...

Is there any particular reason as to why certain texts use ##j## and others ##i## when looking at complex numbers? Maths is a relatively easy subject but at times made difficult with all this mix-up... i tend to use a lot of my time in trying to understand author's language and this is also...

[FONT=times new roman]Statement of the problem : [FONT=times new roman]We have to find what is ##\sqrt{i} + \sqrt{-i}## First Attempt (Euler's Formula) : I use the Euler's formula (see Relevant Equations 1) above which yields ##i = e^{i\frac{\pi}{2}}##. Likewise [FONT=times new roman]##-i =...

I need a to calculate the fresnel reflection ratio of a non dielectric material given the incident angle, the refractive indexes of the incident and interfacing materials and the extinction coefficient of the interfacing material. I need to to this without directly using complex numbers, due to...

Snapshot of Mary L. Boas' Mathematical Physics book So, the marked lines say `If we think of P as the point z = x +iy in the complex plane, we could replace (2.3) by a single equation to describe the motion of P` But, until now I have only learned of representing points in the form (x,y), now...

Hey all, I have a very simple question regarding the quotient of complex values. Consider the function: $$f(a) = \sqrt{\frac{a-1i}{a+1i}}$$ where ##i## is the imaginary unit. When I evaluate f(0) in Mathematica, I get ##f(0) = 1i##, as expected. But if I evaluate at a very small value of ##a##...

In another thread This has me curious about "ordering other than our normal ordering." What does this mean? I take it that "normal ordering (of integers)" is ... 0, 1, 2, 3... Do mathematicians consider alternate orderings like ...0, 2, 1, 3... That doesnt seem to make sense to me, that's...

Hello guys, I am refreshing on complex numbers today; kindly see attached. ok for part (a) this is a circle with centre ##(\sqrt{3}, -1)## with radius =##1## thus, we shall have,

The attempt at a solution: I tried the normal method to find the determinant equal to 2j. I ended up with: 2j = -4yj -2xj -2j -x +y then I tried to see if I had to factorize with j so I didn't turn the j^2 into -1 and ended up with 2 different options: 1) 0= y(-4j-j^2) -x(2j-1) -2j 2)...

Let ##\Omega## here be ##\Omega=\sqrt{-u}##, in which it is not difficult to realize that ##\Omega ## is real if ##u<0##; imaginary, if ##u>0##. Now, suppose further that ##u=(a-b)^2## with ##a<0## and ##b>0## real numbers. Bearing this in mind, I want to demonstrate that ##\Omega## is real. To...

Hello guys, I am able to follow the working...but i needed some clarification. By rounding to the nearest integer...did they mean? ##z=1.2-1.4i## is rounded down to ##z=1-i##? I can see from here they came up with simultaneous equation i.e ##(1-i) + (x+iy) = \dfrac{6}{5} - \dfrac{7i}{5}## to...

First I solved 4+j3, which I squared 4 and 3 to equal 16 and 9 then I added them to get 25 and then I got the square root of 25 = 5. Then I plugged it back in to the equation. [50/(5)(50)+100] x 150 to get 50/350x 150= 1/7(150)= 21.42. I've attached the correct answer.

I'm trying to code Newton Raphson's method for finding zeros. I realize that even if the solution is real, it's possible for guesses to be complex. For example: $$y=\sqrt{x-6}-2$$ While 10 is a valid real root, for any guess less than 6, the result is complex. I tried to run the code allowing...

Here is my attempt(photo below), but somehow the solution in the textbook is z= 2 - (3/2)i, and mine is z=(-3/2) +2i. Can someone please tell me where I am making a mistake? I suppose it's something with x being part of the real part of the 1st complex number and x being part of an imaginary...

How can I solve a system of equations with complex numbers 2z+w=7i zi+w=-1 I have tried substituting z with a+bi and I have tried substituting w=7i-2z but didn't get anything useful. Edit: also, I've tried, multiplying lower eq. with -1 so that I can cancel w but I get stuck with 2z and zi and...

My interest is only on part (a). Wah! been going round circles to try understand why the radius = ##2##. I know that the given sequence is both bounded and monotonic. I can state that its bounded above by ##1## and bounded below by ##0##. Now when it comes to the radius=##2##, i can also say...

The problem is as shown...all steps are pretty easy to follow. I need help on the highlighted part in red. How did they come to; ##z^4+8z^2+16-9z^2=0## or is it by manipulating ##-z^2= 8z^2-9z^2?## trial and error ...

how does capacitors and inductors cause phase difference between current and voltage? how does complex number come into play to explain the relation between phase of current and voltage?

Hello, I'm posting here since what follows is not about homework, but constitutes a personal research which underlies some more general questions. As with the infamous "casus irreducibilis" (i.e. finding the real roots of a cubic function sometimes requires intermediate calculations with...

See the work below: I feel like it that I did it correctly. I feel like I skip a step in my induction. Please point any errors.

I am not sure why criss-cross approach would work here, but it seems to get the answer. What would be the reason why we could use this approach? $$\frac {z-1} {z+1} = ni$$ $$\implies \frac {z-1} {z+1} = \frac {ni} {1}$$ $$\implies {(z-1)} \times 1= {ni} \times {(z+1)}$$

(z-3)3=-8, solve for z. I'm new to complex numbers, so I'm stuck on this basic problem: how do you find all real and non-real solutions in the equality, (z-3)^3=-8? Thanks a bunch.

For part (a), ##z##=##\dfrac {3+i}{3-i}## ⋅##\dfrac {3+i}{3+i}## ##z##=##\dfrac {4}{5}##+##\dfrac {3}{5}i## part (b) no problem as long as one understands the argand plane... For part (c) Modulus of ##z=1## and Modulus of ##z-z^*##=##\frac{6}{5}i##

Find the question below; note that no solution is provided for this question. My approach; Find part of my sketch here; * My diagram may not be accurate..i just noted that, ##OP## takes smallest value of ##12## when ##|z+5|=|z-5|## i.e at the end of its minor axis and greatest value ##13##...

Easy questions, just a lot of computation... $$\frac {z}{z^*}=\frac {a+bi}{a-bi} ×\frac {a+bi}{a+bi}$$ $$c+di=\frac {a^2-b^2}{a^2+b^2}+\frac {2abi}{a^2+b^2}$$ $$⇒c^2= \frac {a^4-2a^2b^2+b^4}{(a^2+b^2)^2}$$ $$⇒d^2= \frac {4a^2b^2}{(a^2+b^2)^2}$$ Therefore...

$$(1+i)z+(2-i)w=3+4i$$ $$iz+(3+i)w=-1+5i$$ ok, multiplying the first equation by##(1-i)## and the second equation by ##i##, we get, $$2z+(1-3i)w=7+i$$ $$-z+(-1+3i)w=-5-i$$ adding the two equations, we get ##z=2##, We know that, $$iz+(3+i)w=-1+5i$$ $$⇒2i+(3+i)w=-1+5i$$...

I went ahead and tried to prove by induction but I got stuck at the base case for ## N =1 ## ( in my course we don't define ## 0 ## as natural so that's why I started from ## N = 1 ## ) which gives ## \sum_{k=0}^1 z_k = 1 + z = 1+ a + ib ## . I need to show that this is equal to ## \frac{1-...

Hi Here is my attempt at a solution for problems 1) and 2) that can be found within the summary. Problem 1) a = 3-2i b= -6-4i c= 4+ 6i d= -4+3i Now, to calculate each vector modulus, I applied the following formula: $$\left| Vector modulus \right| = \sqrt {(a^2 + b^2) }$$ where a = real part...

Hi Guys Finally after a great struggle I have managed to develop and solve my Lagrangian system. I have checked it numerous times over and over and I believe that everything is correct. I have attached a PDF which has the derivation of the system. Please ignore all the forcing functions and...

I need help actually creating the proof. I've done the scratch needed for the problem, it's just forming the proof that I need help in. Bar(a+bi/c+di)= (a-bi) / (c-di) Bar ((a+bi/c+di)*(c-di/c-di)) = ((a-bi/c-di)*(c+di/c+di)) Bar((ac+bd/c^2 +d^2)+(i(bc-ad)/c^2+d^2)) =...

To find √(-2)√(-3). Method 1. √(-2)√(-3) = √( (-2)(-3) ) = √(6). Method 2. √(-2)√(-3) = √( (-1)(2) )√( (-1)(3) ) = √((-1)√(2)√(-1)√(3) = i√(2)i√(3) = (i)(i)√(2)√(3) = -1√( (2)(3) ) =-√6. Why don't the two methods give the same answer? Thanks for any help.

This Mathologer video explains e raised to the i pi in a way that even Homer Simpson can understand.

Hello, I have this (I am solving scholarship exams)math problem and I don't quite know what to do with it , Could You please help? The exercise is about complex numbers and it says: Calculate in the algebraic form(a+bi) I thought on applying substitution since -1=i^2 and z is the real part but...

Hi PF community, I'm reading about complex numbers and i have some questions about the argument of a complex number that i can't solve with Google or reading again the same page. Well, my first doubt is about , i can't understand where come this and why there is some random integer, i...

Assume a transformer as above, with 230V L-N, and I want to work out the L-L voltage. A phasor diagram will show me that the voltages are 120° out of phase. (230∠0°) + (230∠120°) = (230cos0 + j230sin0) + (230cos120 + j230sin120) = 230 + (-115 + j199.2) 115 + j199.2 = 230∠60 What I’m looking...

The impedance Z = R -j/wC + ##\frac{1}{\frac{1}{R} - \frac{\omega C}{j}}## But,1/wC=R So, solving this, I find: Z= 3R/2(1-j) |Z| =##\frac{3R}{\sqrt 2}## I =##\frac{V_i \sqrt 2} {3R}## Vi - IR-IXc =Vo Solving this, ##Vo = V_i -\frac {V_i \sqrt 2}{3} - \frac{V_i \sqrt 2}{3R} \frac{-j}{wC}##...

I think the solution should be: METHOD #1: \begin{align} (\sqrt[4] {-1})^4 & = (-1)^{\frac 4 4} \nonumber \\ & = (-1)^1 \text{, can reduce 4/4 since base is a constant and not a variable in ℝ} \nonumber \\ & = -1 \nonumber \end{align} Alternatively, METHOD #2 for same answer is...

Below is the problem and the correct answer for this algebra problem is 7√2. But I cannot get to the correct answer.