The problem says: Find all solutions z of the equation: z^6 - 4z^3 + 4 = i
First I factored the equation into (z^3 -2)^2 = i, set w= z^3 -2 and solved w^2 = i for w_1 = sqrt(i) and w_2 = -sqrt(i). I tried setting z^3 - 2 = sqrt(i) and solving but I get stuck there. I really have no idea how...
Homework Statement
Draw the set of points in the complex plane satisfying the equation |z + 3i| = 4
Homework Equations
The Attempt at a SolutionI don't know what z is supposed to be. In class, we've been using z to stand for a complex number (x + yi). Am I supposed to substitute...
Homework Statement
z is a complex number.
Find all the solutions of
(z+1)^5 = z^5
The Attempt at a Solution
Of course one could expand (z+1)^5, but I remeber our professor solving this with roots of unity. Can anyone help?
Homework Statement
hi
in trying to solve 4z - 2conj(z) + i = 0
Homework Equations
The Attempt at a Solution
the calculator spits out z = -1/2*i
i did 4(a+bi) - 2(a - bi) + i = 0
4a + 4bi - 2a + 2bi + i = 0
2a + 6bi + i = 0
i get z = -1/6*i
did i mess up...
Homework Statement
Find all complex solutions to the following equation:
3(x^2 + y^2) + (x - iy)^2 + 2(x + iy) = 0 Homework Equations
I want to use the quadratic formula, but not sure if it applies here.
The Attempt at a Solution
This is as far as I can get. What I would like is some idea...
Homework Statement
So you basically have to prove this big long equation d=2v2cos(o)sin(O+o)/gcos(O)
Homework Equations
sin(O+o)=sin(O)cos(o)+sin(o)cos(O)
The Attempt at a Solution
so i have weeded it down to this...
Homework Statement
Let z be a complex number. I want to solve cos(z)= -i*sin(z).
The Attempt at a Solution
Here's my work:
cos(z) = -i*sin(z) implies cos(z) + isin(z) = 0.
Therefore exp(i*z) = 0. Now put z= x+iy then i*z = i*(x+iy) = ix - y, hence
exp(i*z) = exp(ix-y) =...
Homework Statement
A simple way to factor the left hand side of the given equation.
Homework Equations
E^2-E^2(\frac{\frac{F^2vt}{c}}{M})+(\frac{\frac{F^4v^2t^2}{c^2}}{M^2})=-p+(\frac{\frac{F^4v^2t^2}{c^2}}{M^2})
The Attempt at a Solution
I would, just that i am confused on...
Homework Statement
determine the number of roots, counting multiplicities, of the equation z^7-5*z^3+12=0
in side the annulus 1<=|z|<2
Homework Equations use rouche's theorem
The Attempt at a Solution
all i want to do is finding all the solution in a area.
i know the Newton's method, but the problem is how can i divide the area into much smaller areas that i can judge it whether include a solution and can only have one solution.
and how can judge it?
any talk will be appreciated...
Z(x)=(50*m+m^2*tanh((a+b*i)*x))/(m+50*tanh((a+b*i)*x))
where i is imaginary unit, x is the independent variable, Z is the dependent variable, and a, b and m are the parameters.
We measures variable Z at a discrete set of values of variable x:
x=[1, 2, 3, 4, ...20]
Z=[57.6286+0.6328*i...
Homework Statement
I am reading the book Wiley Signal And Systems-Simon Haykin and at the page 34 i read the followings
Consider the complex exponential e^j*theta. Using Euler's identity we may expand this term as e^j*theta=cos(theta)+jsin(theta)
This result indicates that we may express the...
Homework Statement
Find all roots of the equation cos(z)=2 (z is a complex number)Homework Equations
The Attempt at a Solution
What do they mean find the roots of this equation? We're just going over trig functions and it doesn't say anything about roots so I'm not sure what they're...
find all solutions of the given equation: (z+1)^4=1-i
im not sure if i did this right, but here's what i did
the first thing that i did was notice that
1-i = 2^1/2 * (cos (pi/4) + i*sin(pi/4))
then i found
z= [2^(1/2*1/4) * (cos (pi/4) + i*sin (pi/4) )^1/4] -1
then using de moivre's thrm...
Solve the following equations for the real variables x and y.
(b) \left ( \frac{1 + i}{1 - i} \right )^2 + \frac{1}{x + iy} = 1 + i
I reduced this to
2x + 2iy + ix - y - 1 = 0
but I cannot get any further. Have I reduced it correctly?
im not sure i can fully remember the rules for complex numbers but i have to solve an equation that has 3 solutions to equal zero. so far i have (x=2), (x=i) and i think there was a + rule for a solution with complex numbers that there would always be a conjugate solution of it. so i figured...