Your question is unclear, but here is an overview.
Complex numbers are numbers in the form a+bi, where i is the square root of -1.
To add or subtract complex numbers, simply combine like terms.
To multiply, use the good old FOIL method.
To divide by a complex number, multiply both sides of the fraction by the conjugate of the denominator.
Good luck!
2006-08-01 16:36:00
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answer #1
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answered by jenh42002 7
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(m)
Properties of complex numbers
— Function: double gsl_complex_arg (gsl_complex z)
This function returns the argument of the complex number z, \arg(z), where -\pi < \arg(z) <= \pi.
— Function: double gsl_complex_abs (gsl_complex z)
This function returns the magnitude of the complex number z, |z|.
— Function: double gsl_complex_abs2 (gsl_complex z)
This function returns the squared magnitude of the complex number z, |z|^2.
— Function: double gsl_complex_logabs (gsl_complex z)
This function returns the natural logarithm of the magnitude of the complex number z, \log|z|. It allows an accurate evaluation of \log|z| when |z| is close to one. The direct evaluation of log(gsl_complex_abs(z)) would lead to a loss of precision in this case.
2006-08-01 16:38:26
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answer #2
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answered by mallimalar_2000 7
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227
2006-08-01 16:31:30
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answer #3
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answered by wise old,man 3
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rectangular form: P+Qj P is the real part Q is the imaginary part j=sqrt(-1)
Conjugate of P+Qj = P-Qj
R=sqrt(P^2+Q^2): D=atan(Q/P)
Polar form R angle(D) : R is the magnitude D is the angle
Exponential form: R*e^(jD)
e^(j*pi)=j
1/j=-j
Thats about it.
2006-08-01 17:39:14
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answer #4
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answered by DoctaB01 2
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a complex no is, Z= a+bi Zconju.=a-bi
where a is real part & b imaginary i=sq root of -1
prop. z+zconj=2real part
z-zconj=2imaginary part
z.zconj=|z|
|z1+z2|=|z1| +|z2|+2z1z2
2006-08-01 18:09:48
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answer #5
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answered by priya 2
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Z1 = a1 + ib1;
z2 = a2 + ib2;
then some of the properties
1. Z1 + Z2 = Z2 + Z1;
2. Z1 * Z2 = Z2 * Z1;
3. Z1 - Z2 != Z2 - Z1;
4 . Z1 / Z2 != Z2/Z1;
2006-08-01 17:56:12
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answer #6
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answered by yours_ganeshvr 1
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[r (cos θ + i sin θ)]^n = r^n (cos nθ + i sin nθ)
[r (cos θ + i sin θ)]^(1/n) = r^(1/n) (cos (θ + 2kpi)/n + i sin(θ + 2kpi)/n)
where
z = x + yi = r (cos θ + i sin θ)
Called the polar form of complex numbers.
^_^
2006-08-01 22:46:47
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answer #7
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answered by kevin! 5
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rational and irrational....
2006-08-01 16:39:06
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answer #8
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answered by brian b 1
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