What is Complex number: Definition and 437 Discussions
In mathematics, a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a symbol called the imaginary unit, and satisfying the equation i2 = −1. Because no "real" number satisfies this equation, i was called an imaginary number by René Descartes. For the complex number a + bi, a is called the real part and b is called the imaginary part. The set of complex numbers is denoted by either of the symbols
C
{\displaystyle \mathbb {C} }
or C. Despite the historical nomenclature "imaginary", complex numbers are regarded in the mathematical sciences as just as "real" as the real numbers and are fundamental in many aspects of the scientific description of the natural world.Complex numbers allow solutions to all polynomial equations, even those that have no solutions in real numbers. More precisely, the fundamental theorem of algebra asserts that every polynomial equation with real or complex coefficients has a solution which is a complex number. For example, the equation
(
x
+
1
)
2
=
−
9
{\displaystyle (x+1)^{2}=-9}
has no real solution, since the square of a real number cannot be negative, but has the two nonreal complex solutions −1 + 3i and −1 − 3i.
Addition, subtraction and multiplication of complex numbers can be naturally defined by using the rule i2 = −1 combined with the associative, commutative and distributive laws. Every nonzero complex number has a multiplicative inverse. This makes the complex numbers a field that has the real numbers as a subfield. The complex numbers form also a real vector space of dimension two, with {1, i} as a standard basis.
This standard basis makes the complex numbers a Cartesian plane, called the complex plane. This allows a geometric interpretation of the complex numbers and their operations, and conversely expressing in terms of complex numbers some geometric properties and constructions. For example, the real numbers form the real line which is identified to the horizontal axis of the complex plane. The complex numbers of absolute value one form the unit circle. The addition of a complex number is a translation in the complex plane, and the multiplication by a complex number is a similarity centered at the origin. The complex conjugation is the reflection symmetry with respect to the real axis. The complex absolute value is a Euclidean norm.
In summary, the complex numbers form a rich structure that is simultaneously an algebraically closed field, a commutative algebra over the reals, and a Euclidean vector space of dimension two.
This is the question as it appears on the pdf. copy;
##z=2\left[\cos \dfrac{3π}{4} + i \sin \dfrac{3π}{4}\right]##
My approach;
##\dfrac{3π}{4}=135^0##
##\tan 135^0=-\tan 45^0=\dfrac{-\sqrt{2}}{\sqrt{2}}##
therefore,
##z=-\sqrt{2}+\sqrt{2}i##
There may be a better approach.
Hi, PF, so long, I have a naive question: is ##\pi+\arctan{(2)}## a complex number at the second quadrant? To define a single-valued function, the principal argument of ##w## (denoted ##\mbox{Arg (w)}## is unique. This is because it is sometimes convenient to restric ##\theta=\arg{(w)}## to an...
Kindly see attached...I just want to understand why for the case; ##(-1+i)^\frac {1}{3}## they divided by ##3## when working out the angles...
Am assuming they used;
##(\cos x + i \sin x)^n = \cos nx + i \sin nx## and here, we require ##n## to be positive integers...unless I am not getting...
Hello! (Not sure if this is pre or post calc,if I am in the wrong forum feel free to move it)
So im given this complex number ## z = \frac{6}{1-i} ## and I am susposed to get it in polar form as well as z = a+bi
I did that; z = 3+3i and polar form ##z =\sqrt{18} *e^{\pi/4 i} ##
Now Im...
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?
$$\frac {z-1} {z+1} = ni$$
$$\implies \frac {z-1} {z+1} = \frac {ni} {1}$$
$$\implies {(z-1)} \times 1= {ni} \times {(z+1)}$$
I apologize in advance for my English.
I want to know if my solution is correct. :)
To verify that for every complex number z, the numbers z + z¯ and z × z¯ are real.
My solution:
z = a + bi
z¯ = a - bi
z + z¯ = a + bi + a - bi = 2a ∈ R
z × z¯ = (a + bi) × (a - bi) = a^2 + b^2 ∈ R
...and is it ever useful?
An arbitrary complex number has the form ##z = a + bi## where ##a, b \in \mathbb{R}## and the dot product of two arbitrary vectors ##\vec{v} = \binom{v_1}{v_2}## and equivalently for vector ##\vec{w}## is ##\vec{v} \cdot \vec{w} = v_1 w_1 + v_2 w_3## Then the ##z## may...
Let z = x + iy
$$\arg \left(\frac{1+z^2}{1 + \bar z^{2}}\right)=\arg (1+z^2) - \arg (1 + \bar z^{2})$$
$$=\arg (1+x^2+i2xy-y^2)-\arg(1+x^2-i2xy+y^2)$$
Then I stuck.
I also tried:
$$\frac{1+z^2}{1 + \bar z^{2}}=\frac{1+x^2+i2xy-y^2}{1+x^2-i2xy+y^2}$$
But also stuck
How to do this question...
Find the problem here; ( i do not have the solutions...i seek alternative ways of doing the problems)
ok, i let ##z=x+iy## and ##z^*= x-iy##... i ended up with the simultaneous equation;
##2x+y=4##
##x+2y=-1##
##x=1## and ##y=2##
therefore our complex number is ##z=1+2i##
Firstly I converted the given equation to a quadratic equation which is
##z^2- (\sqrt3)z+1=0##
I got two solutions:
1st sol ##z=\frac {(\sqrt3 + i)} {2}##
2nd sol ## z=\frac {(\sqrt3 -1)} {2}##
Then I found modulus and argument for both solution . Modulus=1 Arguments are ##\frac...
This is pretty straightforward,
Let ##z=a+bi##
## \bigl(\Re (z))=a, \bigl(\Im (z))=b##
##zz^*=(a+bi)(a-bi)=a^2+b^2 =\bigl(\Re (z))^2+\bigl(\Im (z))^2##
Any other approach? this are pretty simple questions ...all the same its good to explore different perspective on the same...
My attempt:
Let us put $\frac{1}{i+t} = \frac{1+e^{is}}{2i} \Rightarrow \frac{2i}{i+t} -1= e^{is}$
So, $\cos{s}- i\sin{s}= \frac{2i}{i+t} - 1,\Rightarrow \cos^2{(s)} - \sin^2{(s)} = \frac{-2}{(i+t)^2} +1 -\frac{4i}{i+t}$
After doing some more mathematical computations, I got $\cos{s}=...
Summary:: Hi, I tried attempting this problem in alternating current in order to find out the phasors as complex numbers, and I would be more than grateful if someone could peer review it, and confirm my calculations (Please see below both the Figure and the calculations)
Please find attached...
(e^(i*theta))^2 = (sin(theta)+i*cos(theta))^2 = cos(theta)^2 - sin(theta)^2 + 2*i*sin(theta)*cos(theta), so the real part would be: cos(theta)^2 - sin(theta)^2, and the imaginary part would be: 2*i*sin(theta)*cos(theta). But then I don't know where to start with the modulus or the argument?
1.4.1 Miliani HS
Find all complex numbers x which satisfy the given condition
$\begin{array}{rl}
1+x&=\sqrt{10+2x} \\
(1+x)^2&=10+2x\\
1+2x+x^2&=10+2x\\
x^2-9&=0\\
(x-3)(x+3)&=0
\end{array}$
ok looks these are not complex numbers unless we go back the the...
Dear Everyone,
This post is not a homework assignment...
I want to use the quartic formula. In one step is to solve the resolvent cubic. I know that there is 3 real solutions this particular resolvent cubic. I want to know how Bombelli got his answers before the discovery of the trigonometric...
I have as a solution for part one:
c=(a)/(a^2 + b^2)
d=(-b)/(a^2 + b^2)
Which matches with the solution manual.
The manual goes on to give the solution for part b:
(a+bi) * ( (a)/(a^2 + b^2) - ((b)/(a^2 + b^2))i ) = 1
I'd simply like to know where the 'i' at the end of the second...
This is a quote from "Calculus", by Robert A. Adams. It's a translation from spanish:
"Roots of square numbers
If ##a## is a positive real number, there exist two different real numbers whose square is ##a##. They are
##\sqrt{a}\;## (the positive square root of ##a##)
##-\sqrt{a}\;## (the...
##\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!
Summary:: Hello, my question asks if the complex exponential equation 4e^(-j) is equal to 4 ∠-180°. I tried to use polar/rectangular conversions: a+bj=c∠θ with c=(√a^2 +b^2) and θ=tan^(-1)[b/a]
4e^(-j)=4 ∠-180°
c=4, 4=(√a^2 +b^2)
solving for a : a=(√16-b^2)
θ=tan^(-1)[b/a]= -1
b/(√16-b^2)=...
Consider an equation, $$\tilde{x_0}
= \ln(X+ i\delta),$$ where X may be positive or negative and ##0< \delta \ll 1##. Now, if ##X>0## this evaluates to ##\ln(X)## in some limiting prescription for ##\delta \rightarrow 0## while if ##X<0##, we get ##\ln(-X) + i \pi. ##
Now, consider...
So far I've got the real part and imaginary part of this complex number. Assume: ##z=\sin (x+iy)##, then
1. Real part: ##\sin x \cosh y##
2. Imaginary part: ##\cos x \sinh y##
If I use the absolute value formula, I got ##|z|=\sqrt{\sin^2 {x}.\cosh^2 {y}+\cos^2 {x}.\sinh^2 {y} }##
How to...
Hello! :smile:
I am locked in an exercise.
I must find (and graph) the complex numbers that verify the equation:
##z^2=\bar z^2 ##
If ##z=x+iy## then:
##(x+iy)^2=(x-iy)^2 ##
and operating and simplifying,
##4.x.yi=0 ##
and here I don't know how to continue...
can you help me with ideas?
thanks!
From solving the characteristic equations, I got that ##\lambda = 0.5 \pm 1.5i##. Since using either value yields the same answer, let ##\lambda = 0.5 - 1.5i##. Then from solving the system for the eigenvector, I get that the eigenvector is ##{i}\choose{1.5}##. Hence the complex solution is...
Z can be any point on the argand diagram so if z molous is less than 2 , is that somehow giving us the distance from origin? But how i assumed mod sign only makes things positive therefore its not sqrt( (x+yi)^2 ) = distance ??
I can do question (a). For question (b), I can not see the relation to question (a). Can we really do question (b) using result from (a)? Please give me little hint to relate them
Thanks
Hi
I know that division of a real number by zero is not defined. I just came across the following in a textbook on Complex Analysis by Priestley , " we are allowed to divide a complex number by zero as long as the complex number ≠ 0 "
Is this correct ? What happens if the complex number is...
If I had to guess what the complex matrix would look like, it would be:
##T(x+iy)=(xa-by)+i(ya+bx)=\begin{pmatrix}
a+bi & 0 \\
0 & -b+ai\end{pmatrix}\begin{pmatrix}
x \\
y \end{pmatrix}##
I'm not too sure on where everything goes; it's my first time fiddling with complex numbers, and moreover...
Homework Statement
Find the value of (-√3 + i)43/243
Homework Equations
The Attempt at a Solution
I do not know how to really go about this problem.
I know that i0=1, i1=i, i2=-1, i3=-i, and I tried to use that to help but I got to no where, I also tried to break up the exponent into...
Homework Statement
If ##\text{arg}(w)=\frac{\pi}{4}## and ##|w\cdot \bar{w}|=20##, then what is ##w## of the form ##a+bi##.
Homework Equations
The Attempt at a Solution
The only way for the argument of ##w## to be ##\frac{\pi}{4}## is when ##a+bi## where ##a=b \in \mathbb{Z}## right?
Hello, I need to read a fortran data with complex numbers and real numbers,
the first column is the real numbers, the second and third complex numbers (real, imaginary).
I need to read the first 64 lines and then the next 64 lines in separate ways and save in a variable. for example
read from...
Homework Statement
(1+2i+3i2)/(1-2i+3i2)
answer options : a : 1 b: -i c: i d: 0
Homework Equations
what is the most easy method to solve it ,
The Attempt at a Solution
are they conjugate to each other ? if they are than z/zconjugate =1 ,
but how can...
Homework Statement
[/B]
Homework Equations
The Attempt at a Solution
I had no problems with part a and was able to form the equation of the circle and get its centre/radius.
It's part b that I'm stuck on.
My notes show that for Z < 3 I would shade inside the circle but the mark scheme...
Mentor note: Thread moved from technical section, so missing the homework template.
Hi all, I have a homework problem which asks me to compute the complex number cos(π/4 + π/4 i).
I've been playing around with it for a while now and just can't seem to get the answer I get when using Wolfram...
Homework Statement
If Z= (1)/(z conjugate) then Z : ?
Homework Equations
The Attempt at a Solution
let z= a+bi
the z conjugate= a-bi
(a+bi)=(1)/(a-bi)
(a+bi)(a-bi)=1
a2+b2=1
does it tell from this expresssion that the complex number is a pure real ?
Homework Statement
if z=(x-iy)/(x+iy) then modulus of z is :
Homework Equations
The Attempt at a Solution
(x-iy)/(x+iy)= (x2-y2-2x(iy))/(x2+y2)
i can't get the real part and the imaginary part to take the modulus :
but the answer in any way could be = 1 ?
the answer in the book is 1 .
Homework Statement
Value of x and y , when (x+yi)2= 5+4i
Homework Equations
The Attempt at a Solution
x2+2x(iy)-y2=5+4i
x2-y2=5 -------> (1)
2x(iy)=4i (imaginary part)
xy=2 --------> (2)
solving the two equations
x=2.388 and y=0.838
or x=-2.388 or y=-0.838
is this the right way to...
Homework Statement
Are these functions homomorphisms, determine the kernel and image, and identify the quotient group up to isomorphism?
C^∗ is the group of non-zero complex numbers under multiplication, and C is the group of all complex numbers under addition.
Homework Equations
φ1 : C−→C...
My question is perhaps as much about the philosophy of math as it is about the specific tools of math: is perpendicularity and rotation integral and fundamental to the concept of multiplication - in all number spaces?
As I understand it, the product of complex numbers x = (a, ib) and y = (c...
Hi, I had a question I was working on a while back, and whilst I got the correct answer for it, I was told that there was a second solution to it that I missed.
Here is the question.
]
I worked my answer out to be sqrt(2)(cos(75)+i(sin(75))), however, it appears there is a second solution...