Before running this m-file you need to download the files
http://www.wam.umd.edu/~petersd/463/Arg.m
http://www.wam.umd.edu/~petersd/463/complexpolargridplot.m
We will use I and Pi for the symbolic versions of i and pi:
I = sym('sqrt(-1)'); Pi = sym('pi'); format compact % don't put blank lines between answers
z1 = 1-I; z2 = -2-2*I; t1 = Arg(z1) t2 = Arg(z2)
t1 = -1/4*pi t2 = -3/4*pi
p = z1*z2
Arg(p)
t1 + t2 % Note: t1+h2 is not in (-pi,pi], hence this is not Arg(p)
p = -4 ans = pi ans = -pi
q = z1/z2
Arg(q)
t1 - t2 % Note: t1-t2 is in (-pi,pi], hence this is Arg(q)
q = 1/2*sqrt(-1) ans = 1/2*pi ans = 1/2*pi
z = 1 + I r = abs(z) th = Arg(z) w1 = simplify( r^21*(cos(21*th) + I* sin(21*th)) ) w2 = simplify( r^-5*(cos(-5*th) + I* sin(-5*th)) ) plot(double(z).^(1:8),'o-'); axis equal; grid on
z = 1+sqrt(-1) r = 2^(1/2) th = 1/4*pi w1 = -1024-1024*sqrt(-1) w2 = -1/8+1/8*sqrt(-1)
w = -1/4 - 1/4*I; t = Arg(w) r = abs(w) z1 = simplify( r^(1/3)*(cos(t/3) + I*sin(t/3)) ) % principal value omega = cos(2*Pi/3) + I*sin(2*Pi/3) % root of 1 z2 = simplify(z1*omega) z3 = simplify(z1*omega^2) plot( double([z1 z2 z3 z1]),'o-'); axis equal; grid on
t = -3/4*pi r = 1/4*2^(1/2) z1 = 1/2-1/2*sqrt(-1) omega = -1/2+1/2*sqrt(-1)*3^(1/2) z2 = 1/4*sqrt(-1)*3^(1/2)-1/4+1/4*sqrt(-1)+1/4*3^(1/2) z3 = -1/4*sqrt(-1)*3^(1/2)-1/4+1/4*sqrt(-1)-1/4*3^(1/2)
w = -8 + 8*sym('sqrt(3)')*I; t = Arg(w) r = abs(w) z1 = simplify( r^(1/4)*(cos(t/4) + I*sin(t/4)) ) % principal value omega = cos(2*Pi/4) + I*sin(2*Pi/4) % root of 1 z2 = simplify(z1*omega) z3 = simplify(z1*omega^2) z4 = simplify(z1*omega^3) plot( double([z1 z2 z3 z4 z1]),'o-'); axis equal; grid on
t = 2/3*pi r = 16 z1 = 3^(1/2)+sqrt(-1) omega = sqrt(-1) z2 = sqrt(-1)*(3^(1/2)+sqrt(-1)) z3 = -3^(1/2)-sqrt(-1) z4 = -sqrt(-1)*3^(1/2)+1
% x + i y = (u + i v)^2 = (u^2 - v^2) + i (2uv) % x = u^2 - v^2 (eq1) % y = 2uv (eq2) % (x + y)^2 = (u^2 - v^2)^2 + (2uv)^2 = (u^2 + v^2)^2 % r = u^2 + v^2 (eq3) % (eq1) and (eq3) give % u^2 = (r + x)/2 % v^2 = (r - x)/2 % By definition of principal value of root we must have u >= 0 % (eq2) and u>=0 imply that sign(v) = sign(y) % RESULT: u = sqrt((r+x)/2) % v = sign(y)*sqrt((r-x)/2)
x = sym(3); y = sym(-4); r = sqrt(x^2+y^2) u = sqrt((r+x)/2) v = -sqrt((r-x)/2) % "-" since y<0 square = (u + I*v)^2 % check result root = sqrt(3-4*I) % The sqrt command gives the same result
r = 5 u = 2 v = -1 square = 3-4*sqrt(-1) root = 2-sqrt(-1)
x = sym(4); y = sym(3); r = sqrt(x^2+y^2) u = sqrt((r+x)/2) v = sqrt((r-x)/2) % "+" since y<0 square = simplify( (u + I*v)^2 ) % check result root = sqrt(4+3*I) % The sqrt command gives the same result
r = 5 u = 3/2*2^(1/2) v = 1/2*2^(1/2) square = 4+3*sqrt(-1) root = 3/2*2^(1/2)+1/2*sqrt(-1)*2^(1/2)
a = 1; b = 1 - I; c = 4 + 7*I;
d = b^2-4*a*c
s = sqrt(d) % use formula from problem (4) for finding root
z1 = (-b+s)/(2*a)
z2 = (-b-s)/(2*a)
d = -16-30*sqrt(-1) s = 3-5*sqrt(-1) z1 = 1-2*sqrt(-1) z2 = -2+3*sqrt(-1)
Let w:=z^2, then we have to solve the quadratic equation w^2 - 6w + 25 = 0:
a=sym(1); b=sym(-6); c=sym(25); s = sqrt(b^2-4*a*c) w1 = (-b+s)/(2*a) w2 = (-b-s)/(2*a) z1 = sqrt(w1) % use formula from problem (4) for finding root z2 = -z1 z3 = sqrt(w2) % use formula from problem (4) for finding root z4 = -z3
s = 8*sqrt(-1) w1 = 3+4*sqrt(-1) w2 = 3-4*sqrt(-1) z1 = 2+sqrt(-1) z2 = -2-sqrt(-1) z3 = 2-sqrt(-1) z4 = -2+sqrt(-1)
% Assume that % % pvsqrt(z^2) = z (1) % % As Arg(pvsqrt(w)) is always in (-pi/2,pi/2] we must have % % Arg(z) in (-pi/2,pi/2] (2) % % Let t=Art(z), r=|z| and assume that (2) holds. Then z^2 = r^2 exp(i*2*t) % (2) implies that 2*t is in (-pi,pi], hence Arg(z^2) = 2*t and % pvsqrt( z^2 ) = sqrt(r^2) exp(i*(2*t)/2) = r exp(i*t) = z. % % ANSWER: % in polar coordinates: all numbers of the form r*exp(i*t) where % r>=0, t in (-pi/2,pi/2] % in cartesian coordinates: all numbers of the form x + i*y where % x>0 or (x=0 and y>=0)
complexpolargridplot('z',1,2,0,pi/2)
complexpolargridplot('z^3',1,2,0,pi/2)
complexpolargridplot('z^-1',1,2,0,pi/2)
complexpolargridplot('sqrt(z)',1,2,0,pi/2)
