Asked by jack
a) Obtain all solutions of the equation z^3 +1 = 0
b) Let z = x + iy. Obtain the real and imaginary parts of the function
f(z) = 1/1+z
c) Let f(x + iy) = x^2 - y^2 + iv( x,y). Determine a function v such
that f is differentiable in the whole complex plane. Express f
as a function of a single complex variable z.
d) Let f(x + iy) = u(x,y) + ic be analytic, where c is a constant.
Show that u must also be a constant.
b) Let z = x + iy. Obtain the real and imaginary parts of the function
f(z) = 1/1+z
c) Let f(x + iy) = x^2 - y^2 + iv( x,y). Determine a function v such
that f is differentiable in the whole complex plane. Express f
as a function of a single complex variable z.
d) Let f(x + iy) = u(x,y) + ic be analytic, where c is a constant.
Show that u must also be a constant.
Answers
Answered by
Damon
T is theta
z = r e^iT = r (cos T + i sin T)
z^3 = r^3 e^3iT = -1
if r = 1
e^3 i T = -1 = cos 3T + i sin 3T
3T can be pi or 3 pi or 5 pi
so
T = pi/3 or pi or 5 pi/3
so
z = e^i(pi/3) or e^i(pi) or e^i(5 pi/3)
for pi/3
z = cos pi/3 + i sin pi/3
= .5 + .866 i
for pi
z = -1 (well we knew that)
for 5 pi/3
z = .5 - ..866 i I hope :)
z = r e^iT = r (cos T + i sin T)
z^3 = r^3 e^3iT = -1
if r = 1
e^3 i T = -1 = cos 3T + i sin 3T
3T can be pi or 3 pi or 5 pi
so
T = pi/3 or pi or 5 pi/3
so
z = e^i(pi/3) or e^i(pi) or e^i(5 pi/3)
for pi/3
z = cos pi/3 + i sin pi/3
= .5 + .866 i
for pi
z = -1 (well we knew that)
for 5 pi/3
z = .5 - ..866 i I hope :)
Answered by
Damon
b) Let z = x + iy. Obtain the real and imaginary parts of the function
f(z) = 1/(1+z) I assume
f(z) = 1/[(1+x) +iy ]
= [(1+x)-iy] / {[(1+x) +iy ][[(1+x)-iy] }
= [(1+x)-iy] /{(1+x)^2 +y^2}
= (1+x)/{(1+x)^2 +y^2}
- i y /{(1+x)^2 +y^2}]
f(z) = 1/(1+z) I assume
f(z) = 1/[(1+x) +iy ]
= [(1+x)-iy] / {[(1+x) +iy ][[(1+x)-iy] }
= [(1+x)-iy] /{(1+x)^2 +y^2}
= (1+x)/{(1+x)^2 +y^2}
- i y /{(1+x)^2 +y^2}]
Answered by
Damon
That is all I can try tonight.
Answered by
jack
Thank you. Just wondering why 3T can be 3 pi, 5 pi, pi.
Also just need c and d.
Also just need c and d.
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