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Complex Analysis
Q: Evaluate the following integral using residue theorem
∫ coth z / (z-i) dz ; C : |z| = 2
∫ coth z / (z-i) dz ; C : |z| = 2
Complex Analysis
show that the function f(z)=1/a+z/a^2+.....can be continued analytically
Complex Analysis
evaluate close integrals :
closed integral at c 1/(z^4+1) dz with contour c:|z|=4
closed integral c 1/(z^4+10z+9) dz with c:|z|=2
closed integral c e^z/(z^4+5z^3) dz with c:|z|=2
closed integral at c 1/(z^4+1) dz with contour c:|z|=4
closed integral c 1/(z^4+10z+9) dz with c:|z|=2
closed integral c e^z/(z^4+5z^3) dz with c:|z|=2
Complex Analysis
Let w be a complex number, z a 4th root of w.
1) Show that z(k) = p^1/4[cos((θ + 2kπ)/4) + isin((θ + 12kπ)/4)], k = 0, 1, 2, 3, is a formula for the 4th roots of w, where θ is the argument of w and p its modulus.
2) hence , determine 4th roots of 16.
please assist.
1) Show that z(k) = p^1/4[cos((θ + 2kπ)/4) + isin((θ + 12kπ)/4)], k = 0, 1, 2, 3, is a formula for the 4th roots of w, where θ is the argument of w and p its modulus.
2) hence , determine 4th roots of 16.
please assist.
Complex Analysis
Using De Moivre's theorem or otherwise
find the six roots of the equation z^6 +1 = 0
giving your answers in the form e^iw where w= angle teta
find the six roots of the equation z^6 +1 = 0
giving your answers in the form e^iw where w= angle teta
Complex Analysis
Find the original function without finding the corresponding conjugate
u(x,y)=e^xcosy
u(x,y)=e^xcosy
Complex Analysis
Find the original function without finding the corresponding conjugate
u(x,y)=x/x^2+y^2
u(x,y)=x/x^2+y^2
Complex Analysis
Prove that u(x,y) given by the following is harmonic obtain it's corresponding conjugate and original function f(z)
u(x,y)=e^xCosy
u(x,y)=e^xCosy
Complex Analysis
Prove that u(x,y) given by the following is harmonic obtain it's corresponding conjugate and original function f(z)
u(x,y)=x^2-y^2
u(x,y)=x^2-y^2
Complex Analysis
Find the image of the unit circle |z| = 2 under the linear fractional transformation (z + 2)
T(z)= (z−1).
T(z)= (z−1).
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#339914 on Dec 2023
#339914 on Dec 2023