Advanced Engineering Mathematics, 4th Edition
(2020) – Potter – Solutions Manual PDF
complex number - answer-a+bi
Polar Form - answer-z = r(cosθ + isinθ)
r is the - answer-Modulus
θ - answer-Argument
Eurler's Formula - answer-e^iy = cosy+ jsiny
Prouct - answer-zz(bar) = x^2 + y^2
Substraction - answer-Im(z) = 1/2i (z-z(bar))
Addition - answer-Re(z) = 1/2 (z + z(bar))
^n sqr (r angle θ )= - answer-(r angle θ )^1/n = r^1\n [1/n(θ+360)]
Logarithm of complex numbers - answer-In (a+ib) = ln( re^jb) = lnr + jθ
1
, Trigonometric
Cos θ - answer-(e^jθ + e^-jθ) / 2
Trigonometric
sin θ - answer-(e^jθ - e^-jθ) / 2j
Hyperbolic
cosh θ - answer-(e^θ + e^-θ) / 2
Hyperbolic
sinh θ - answer-(e^θ - e^-θ) / 2
cos(iy) - answer-coshy
cosh(iy) - answer-cosy
Sin(iy) - answer-isinh(y)
sinh(iy) - answer-isiny
2
(2020) – Potter – Solutions Manual PDF
complex number - answer-a+bi
Polar Form - answer-z = r(cosθ + isinθ)
r is the - answer-Modulus
θ - answer-Argument
Eurler's Formula - answer-e^iy = cosy+ jsiny
Prouct - answer-zz(bar) = x^2 + y^2
Substraction - answer-Im(z) = 1/2i (z-z(bar))
Addition - answer-Re(z) = 1/2 (z + z(bar))
^n sqr (r angle θ )= - answer-(r angle θ )^1/n = r^1\n [1/n(θ+360)]
Logarithm of complex numbers - answer-In (a+ib) = ln( re^jb) = lnr + jθ
1
, Trigonometric
Cos θ - answer-(e^jθ + e^-jθ) / 2
Trigonometric
sin θ - answer-(e^jθ - e^-jθ) / 2j
Hyperbolic
cosh θ - answer-(e^θ + e^-θ) / 2
Hyperbolic
sinh θ - answer-(e^θ - e^-θ) / 2
cos(iy) - answer-coshy
cosh(iy) - answer-cosy
Sin(iy) - answer-isinh(y)
sinh(iy) - answer-isiny
2
