4
votes
2answers
80 views

Can't solve equation having complex coefficients

Why can't Mathematica solve $\quad\quad\frac{1-i}{\sqrt{2}}=e^{i \alpha } \tan \left(\frac{\beta }{2}\right)$ with the restrictions $\alpha \in [0, 2 \, \pi)$ and $\beta \in [0,\pi]$: ...
1
vote
4answers
161 views

How to extract only real solutions from the result of Solve

For Example, I have the following polynomial equations: $$ \left\{ \begin{array}{ll} \dfrac{2025 (208 x+5 (y-3446))}{52 (y+90)^2}+\dfrac{300 (8 x-21 y+1920) (y-80)}{7 (x+30)^3}+\frac{300 (80 x-21 ...
1
vote
1answer
87 views

How do I solve for the real and complex parts of an equation simultaneously?

If I were to have an equation, say something similar to... (1-Sqrt[x - I y])/(1+Sqrt[x - I y]) = A + I B Where I = Sqrt[-1], is there a way for Mathematica to ...
3
votes
1answer
275 views

HarmonicNumber problem

I am looking for all the roots of HarmonicNumber, in the domain [-30, 1] and [0, 6000] for the real and imaginary parts, respectively, and where the parameter ...
-2
votes
1answer
148 views

How to solve a non-linear equation?

I am trying to solve the following non-linear equation: $\frac{i e^{-x^2} \sqrt{\pi } x}{K^2}+\left(-0.131251-0.0379031 x^2-0.0151154 x^4-0.0100462 x^6+0.5 \left(0.00273859 +0.0400003 K^2\right) ...
0
votes
1answer
82 views

A question about using Reduce [closed]

Can someone help decipher what is the meaning of this computation that Mathematica has done? ...
2
votes
1answer
85 views

Solving Complex Equation Over Reals

I want to solve the equation $\frac{abi}{a+bi}=4-2i$, where $a$ and $b$ are real numbers. I know from hand-solving the answer is $a=5$, $b=-10$. How do I get Mathematica to tell me this? I tried: ...
3
votes
1answer
171 views

When does the real part of Zeta vanish on the critical line?

This seems to be a quite a simple problem but I cannot make it work. I am trying to find all values within a given range for which the real part of the Zeta function vanishes on the line: ...
0
votes
0answers
326 views

Solve equations real and imaginary part separately

For my system of equations, the procedure described in Solving complex equations of using Reduce works no more. How can I separate the real and imaginary part of ...
1
vote
1answer
292 views

Stop Mathematica from giving imaginary solutions

I have the following equation: $$D=\frac{1}{64} \pi A ^3 B \sin \left(C\right)-\frac{1}{2} \pi A B \sin \left(C\right)$$ which I want to solve for $A$. The equation is cubic in $A$ so this should ...
0
votes
1answer
1k views

How can I get the solution of complicated implicit function?

The question is what is the method to solve the implicit function has real and imaginary number. For example, The function is $$F(x,y)=(-I*x + 2*y^2)^2 + x^2 - 4*y^4*Sqrt[1 - I*x/(y^2)]$$ Although ...
1
vote
2answers
821 views

Solving cubic equation for real roots

I'm looking to solve the following cubic equation for x: $\beta\, x^3 - \gamma \,x = c$. I have plugged in some sample values ($\beta = 2$, $\gamma = 5$ and $c = 2$). When I try to solve this ...
2
votes
1answer
2k views

Forcing FindRoot to return only real solutions

FindRoot documentation reports that if the equation and the initial point are reals, the solutions are searched in the real domain. However, in the following case I ...
0
votes
1answer
163 views

Manipulate[]'ing complex roots of an equation using a 2D slider [closed]

I want to make a demonstration of how the complex roots of a polynomial change when I alter the coefficients. Here is my attempt: ...
3
votes
2answers
434 views

Is there a way to solve the Apollonius Circle problem in Mathematica?

Assuming x, a, and b are complex numbers, is there a way to reduce the equation Abs[x - a] == k Abs[x - b] to something like ...
4
votes
4answers
2k views

Solving complex equations

I feel like I am missing a basic, but key point when using Mathematica's Solve or Reduce. ...
9
votes
2answers
975 views

RootSearch for complex or multiple equations

First the background. I'm trying to solve for the roots of a rather messy complex equation. This is not the exact equation, but it's a decent (simpler) stand in: ...