MHB How to Avoid Extraneous Solutions in Solving Complex Equations

[Yankel]

Hello all,

Please look at the following:

Solve the equation:

\[\left | z \right |i+2z=\sqrt{3}\]

where z is a complex number.

I tried solving it, and did the following, which is for some reason wrong. I saw a correct solution. My question to you is why mine is not, i.e., where is my mistake ?

\[i\sqrt{x^{2}+y^{2}}+(2x+2iy)=\sqrt{3}\]

\[(x^{2}+y^{2})(-1)+(4x^{2}+8xiy-4y^{2})=3\]

\[3x^{2}-5y^{2}+8xiy=3\]

\[(1,0),(-1,0)\]

This is definitely wrong. Can you please tell me where my mistake it ?

Thank you !

The correct answer should be: \[\frac{\sqrt{3}}{2}-\frac{1}{2}i\]

 

[I like Serena]

I tried solving it, and did the following, which is for some reason wrong. I saw a correct solution. My question to you is why mine is not, i.e., where is my mistake ?

\[i\sqrt{x^{2}+y^{2}}+(2x+2iy)=\sqrt{3}\]

\[(x^{2}+y^{2})(-1)+(4x^{2}+8xiy-4y^{2})=3\]

Hey Yankel,

You've squared the equation.
However, the left side was not squared correctly.
Note that $(a+b)^2 \ne a^2+b^2$.

Instead, there is no need to square at all.
We can rearrange the equation as:
\[2x + i \left(\sqrt{x^{2}+y^{2}}+2y\right)=\sqrt{3}\]
If follows directly that $x=\frac{\sqrt 3}2$, after which we can solve for the imaginary part to be zero.

 

[Yankel]

Thank you ! Silly mistake (Doh)

Solving your way, I get two solutions (y=1/2 and y=-1/2). One is incorrect. How can I know to ignore it without checking if the equation is valid with each solution ?

 

[I like Serena]

Solving your way, I get two solutions (y=1/2 and y=-1/2). One is incorrect. How can I know to ignore it without checking if the equation is valid with each solution ?

You would have squared to solve the imaginary part to be zero.
That introduces an extraneous solution.
Check just before squaring whether y is supposed to be positive or negative. Then we can tell after (or during) solving which one to discard.