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Complex Numbers
The square root of negative one, which we now call the imaginary unit, was once thought to be an absurd notion. However, through the diligent studies of open-minded mathematicians, it was shown that the real numbers were actually just one part of a larger set of numbers known as the complex numbers, the other part being the imaginary numbers. Please direct all questions about these surprisingly useful and applicable numbers into this category.
887 Questions
What is mode of z in complex no?
In complex mode functions, modules, and procedures cannot operate. For a complex number z = x + yi, first define the absolute value. This would be |z| and is the distance from z to 0 in the complex plane.
Where did the imaginary numbers come from?
Imaginary numbers are numbers whose square is a negative number. They arose as a means of working with square roots of negative numbers; in fact, the first known mention of a square root of a negative number is a very brief one from a work called Stereometrica. It was written in the 1st century CE by a Greek mathematician, Heron of Alexandria. Imaginary (and thus, complex) numbers were not ever accepted widely, though, until the 1700s, because of the work of Euler and Gauss.
Which is the largest number to have a name (example - 100 Hundred)?
It is infinity which can't be exactly determined.
How do you divide complex numbers?
When dividing complex numbers you must:
- Write the problem in fractional form
- Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.
You must remember that a complex number times its conjugate will give a real number.
a complex number 2+2i. the conjugate to this is 2-i1. Multiply both together gives a real number.
(2+2i)(2-2i) = 4 -4i + 4i + (-4i2) (and as i2 = -1) = 8
To divide a complex number by a real number simply divide the real parts by the divisor.
(8+4i)/2 = (4+2i)
To divide a real number by a complex number.
1. make a fraction of the expression 8/(2+2i)
2. multiply by 1. express 1 as a fraction of the divisor's conjunction. 8/(2+2i)*(2-2i)/(2-2i)
3. multiply numerator by numerator and denominator by denominator.
(16-16i)/8
4. and simplify 2-2i
What is the sum of complex numbers?
In (a+bi) + (c+di), you add the real parts using the laws for real numbers and do the same for the imanginary parts.
(a+c)+(b+d)i
Are imaginary and complex numbers the same?
No. A complex number is a number that has both a real part and an imaginary part. Technically,
a pure imaginary number ... which has no real part ... is not a complex number.
How do you solve complex numbers?
That depends what is the problem given, and what you want to solve. You may want to read an introductory article on complex numbers, to learn how you add them, multiply them, etc.
That depends what is the problem given, and what you want to solve. You may want to read an introductory article on complex numbers, to learn how you add them, multiply them, etc.
That depends what is the problem given, and what you want to solve. You may want to read an introductory article on complex numbers, to learn how you add them, multiply them, etc.
That depends what is the problem given, and what you want to solve. You may want to read an introductory article on complex numbers, to learn how you add them, multiply them, etc.
How does the job of a construction worker relates to math?
Relates in many different ways you are constantly working with fractions, there are different formulas for different things like figuring out the rise and run on a set of steps or the rafters to frame a roof.
Is The sum of two complex numbers is sometimes a real number.?
Sure, if the imaginary part is opposite. For example:(3 + 2i) + (5 - 2i) = 8
What is the area of a circle if a point on the circle is 3 plus 2i?
distance from point (0,0) to 3+2i is |3+2i| =sqrt(9+4)=sqrt(13)
so the radius of the circle is sqrt(13) and its area is sqrt(13)*sqrt(13)*Pi = 13*Pi = 40.84
What is the real number of 7 over 8 and 9 over 10?
It depends:
7/8/9/10 can be:
7/(8/(9/10)) = 7/(80/9) = 63/80, or
7/((8/9)/10) = 7/(8/90) = 7/(4/45) = 315/4
(7/8)/(9/10) = 70/72 = 35/36
(7/(8/9))/10 = (63/8)/10 = 63/80
((7/8)/9)/10 = (7/72)/10 = 7/720
* * * * *
There is an "and" between 8 and 9 so the question, in fact, refers to 7/8 + 9/10 which is not ambiguous.
7/8 + 9/10 = 35/40 + 36/40 = 71/40 = 1 and 31/40
What set of number does -1 belong?
To any set that contains it!
It belongs to {-1},
or {45, sqrt(2), -1, pi, -3/7},
or all whole numbers between -43 and 53,
or square roots of 1,
or negatives of counting numbers,
or integers,
or rational numbers,
or real numbers,
or complex numbers,
etc.
What is negative i raised to the i power where i is the imaginary unit equal to square root of -1?
-√(-1)√(-1) can be shortened to -ii, where i = √(-1).
The answer to this question is not trivial. To answer it, we must invoke Euler's formula, which is eix = cos(x) + isin(x).
Substituting in π/2 for x gives us eiπ/2 = cos(π/2) + isin(π/2).
Well, the cos(π/2) is 0, and the sin(π/2) is 1, so the above becomes:
eiπ/2 = i.
So, now we raise both sides to the power of i:
ei*iπ/2 = ii.
Using the basic identity for i:
i*i = i2 = -1.
So, now we have e-π/2 = ii, or -e-π/2 = -ii.
Well, -e-π/2 is a real number, and its value is -0.20788.
For the similar case of (-i)i, we substitute -π/2 into Euler's formula.
This gives us e-iπ/2 = cos(-π/2) + isin(-π/2) = 0 - i = -i.
Once again, raising both sides to the power of i and using the same identity, we get eπ/2 = (-i)i.
Again, the result is a real number, eπ/2, whose value is 4.81047.
--
NOTE.
I'd like to add that (-i)ior ii are not (well-)defined, so there is no correct/definite answer to this question.
Indeed, Euler's formula is valid, but the formula is inclomplete; in fact it is
ei.(x+2.k.π) = cos(x) + i.sin(x) for any integer k.
Therefore,
ei.π.(2.k+1/2) = i for any integer k,
and so
e-π.(2.k+1/2) = ii for any integer k,
proving that it is not well-defined.
Other possible values for ii are e-π.(-2+1/2) =eπ.(3/2) = 111.31777... and
e-π.(2+1/2) =e-π.(5/2) = 3.88203203... x 10-4. The reason for this is, that in order to compute ii we like to write ii=ei.log(i), but to compute log(z) for a complex number z, one has to decide on what branch of log(.), the inverse of the periodic exp(.), z is located. Note that this aspect was ignored in the previous answer. Therefore, there is no correct answer for this question: any branche of log(.) can be chosen to produce an answer that may suit the context. The question does not give this context.
This is an interesting question. Looking at complex numbers graphically, zero is at the intersection of the real and imaginary axis, so it is 0 + 0i. But if you square zero, you get zero, which is not a negative number (a pure imaginary, when squared will give a real negative number), so I'd have to say it is not imaginary.
Why is the log of 0 equal to -infinity why not infinity?
First let's say log100=x
If we rearrange this into exponential form, we get 10x=0
Now we know that a number to ANY exponent will NEVER give 0, it's mathematically impossible, unless you start dealing with infinities.
If we put in infinity for "x", we have 10infinity=0, however, we know this can't be possible, because if 10 is to the power of a positive number (even if it is infinitely small ("x" will just keep getting closer to 0 which will make the answer equal 1, and that's the absolute lowest number you can have with a positive exponent)) then it will never come close to 0.
Instead, if we have 10-infinity=0 we understand that the number becomes infinitely small (closer and closer to 0). Although it never technically reaches 0, it becomes so small that it is negligible. This idea is the same as a limit in calculus.
