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Test the series for convergence : Details

1) [i^n ] / [2^(n+2)]

2) [ n! ^ 2] / [e^n]

3) [1] / [ {square root of (i+n) }^n ]

4) conjugate [ (1 / (n^i ) ]
Another answers :Test the series for convergence : Details

1 ) [3i+n] / [n^3+n+1]

2) [n+i] / [4^n]

3) [ 1/ ( n^i ) ] ^2

4) e ^ ( i cosh n)
Test the series for convergence : Details

1 ) [3i+n] / [n^3+n+1]

2) [n+i] / [4^n]

3) [ 1/ ( n^i ) ] ^2

4) e ^ ( i cosh n)
Test the series for convergence : Details

1) [i^n ] / [2^(n+2)]

2) [ n! ^ 2] / [e^n]

3) [1] / [ {square root of (i+n) }^n ]

4) conjugate [ (1 / (n^i ) ]
Find if these sequence convergent of divergent ( details )

1) [ (1+i)^(1/n) ] / square root(n-1)
2) [2^n! ] / [2^ (n+1)! ]
3) sin ((1+i)/n)
Find if these sequence convergent of divergent ( details )
1) Zn = [ i.(z^n) - n.(3^(n+1)) ] / [ i.n .(2^(n-1)) ]

2) [ conjugate ( 4 n^2- i n +1 ) ] / [ (i n +3) ^2 ]

3) [ 8 ^ (n+1) - 5^(n) ] / [ 5 . (8^n)+ 3 ^(n+1)]
Test series for convergence:

1) conjugate [(1/n^i)]
2) (3i+n) / ( n^3+n+1)
3) [1/(n^i)]^2
4) e^(i cosh n)
Test series for convergence:
1) [ i^n/(2^(n+2)) ]
2) (n!)^2/e^n
3) 1/[root(i+n)]^n
4) e^(i coshn)
5) [n+i/(4^n)]
Find the integral

integral from 0 to 2 pi for

e^(e^(i theta )) d theta

where C: \Z\ =1 Oriented clockwise
Find the integral

integral Curve f(z) = exp^(Z*) dz,
where C is line segment from 4i to 2
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