$$z^2=3+4i ,(z=x+iy)$$ Seems easy? That's what I thought!  I get a system of equations that I can't solve: $x^2-y^2=3$ and $2xiy=4i$ which I then get $x=\frac{2}{y}$  I can't solve the system of equations?
11095
If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?
30520
I haven't found any useful method to solve the following problem: Prove that if $x,y,z\in\mathbb{Z}$ and $x^3+y^3=3z^3$ then $xyz=0$.  Source:
24164
If the equation is $x + y + z = 24$ then it is solvable with stars and bars theorem.  But what to do if it is $3x + y + z = 24$?
38537
Solve for $z$: $$z^2+|z|=0$$ I found this question on my textbook but I unable to derive an answer to it as a modulus of $z$ is present. I am unable to take an approach please guide me ..thanks
98209
The equation $X^{n}  + Y^{n}  = Z^{n}$    , where $n   \geq  3$     is a natural number, has no solutions at all where $X,Y,Z$     are integers.   My solution:  False. Because if we let $X=0  ,Y=0$,  then $0^{3}+0^{3}=0^{3}$  . Thus there is a solution for the equation $X^{n}  + Y^{n}  = Z^{n}$,   where $X,Y,Z$   are integers.  Can anyone please give feedback on my answer and state whether it is correct on not.
75390
Recently I asked a question regarding the diophantine equation $x^2+y^2=z^n$ for $x, y, z, n \in \mathbb{N}$, which to my surprise was answered with the help complex numbers. I find it fascinating that for a question which only concerns integers, and whose answers can only be integers, such an elegant solution comes from the seemingly unrelated complex numbers - looking only at the question and solution one would never suspect that complex numbers were lurking behind the curtain!  Can anyone give some more examples where a problem which seems to deal entirely with real numbers can be solved using complex numbers behind the scenes? One other example which springs to mind for me is solving a homogeneous second order differential equation whose coefficients form a quadratic with complex roots, which in some cases gives real solutions for real coefficients but requires complex arithmetic to calculate.  (If anyone is interested, the original question I asked can be found here: )  EDIT:  I just wanted to thank everyone for all the great answers! I'm working my way through all of them, although some are beyond me for now!
49158
Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.   What are all integral solutions of this equation?   I can only find $x=y=z=1$.
124153
I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon  inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it.   From the definition of the factorial function we know $x! = x(x-1)(x-2)...(2)(1)$  So we can do something like this:  $$ [x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)]$$   we can then factor all of the common terms out on the LHS.  $$ [...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)]$$ and divide the common terms out of the right hand side  $$[x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...]$$  I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution).   If anybody can provide a hint as to how to proceed I would appreciate it.
105240
Describe the set of points on the complex plane for which $|z-2| + |z+2|=4$... So, I know you can solve this instantly, just by using definition, but I want to do it the long way..  So,  $$|x- i*y -2|+ |x-i*y +2| = 4$$       $$(x-2)^2 + y^2 + (x+2)^2 +y^2 =16$$      $$ x^2-4*x+4+y^2+x^2+4*x+ 4 +y^2=16$$  now, $x^2$ and $y^2$ were supposed to eliminate but they don't ...where am I wrong?
58995
Find all possible solutions of $$2^x + 3^y = z^2.$$    My approach.   First I substituted $x = 0$, and got the solution, then for $y = 0$. And for $x &gt;  0$ and $y &gt; 0$ , I just know the Pythagorean triple and got the solution as $(4,2,5)$ and $(4,2,-5)$. Please help me to solve it properly.
1698
Are there any rational numbers $x$ and $y$ such that $x^2+y^2=3$. I think there are no rational solutions, but I haven't been able to prove it.
113894
I am unfamiliar with this type of problem. How does one solve this and under what category of math does this fall under.  Find the integral solutions for $x^2+y^2+z^2=x^2y^2$
76358
How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y&gt;0$ ? ($x,y$ are integers.)
79834
if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$  putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$  i wan,t be able to  go further,could some help me with this
90756
Solve for $x,y,z\in\mathbb{N}$  $\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$  I tried by some general methods but they didn't help me.
4916
Question:  let $x,y,z$ be postive integers and  $x\neq y\neq z$ such $$xyz|(xy-1)(xz-1)(yz-1)$$ Find all $x,y,z$,  my try: $$(xy-1)(yz-1)(xz-1)=x^2y^2z^2-xyz(x+y+z)+(xy+yz+xz)-1$$ then  $$\dfrac{(xy-1)(yz-1)(xz-1)}{xyz}=xyz-(x+y+z)+\dfrac{xy+zy+xz-1}{xyz}$$ so $$\dfrac{xy+yz+xz-1}{xyz}\in N$$ Then I can't get any further  I found $(x,y,z)=(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2)$such it, I don't know/have any other,  also How to prove it.  Thank you very much
85946
For the complex polynomial $z^3 -5z^2 +(7-2i)z  +6i-3 = 0 $   $1)$ show that $2+i $  is a root.  $2)$ solve the given equation.  Attemp to solve: I'm not really sure how to solve this, but I considered using Vietes formulas to comstruct a system of equations with two unknowns, and from there to cacluate the roots.....I'm not sure how to do No. 1 though...  help with this would be appricated...