MHB How to Avoid Extraneous Solutions in Solving Complex Equations

AI Thread Summary
The discussion focuses on solving the complex equation |z|i + 2z = √3, where the original poster made an error by incorrectly squaring the equation, leading to extraneous solutions. A key point raised is that squaring can introduce solutions that do not satisfy the original equation, emphasizing the importance of checking conditions before squaring. The correct approach involves rearranging the equation to isolate the real and imaginary parts without squaring. The conversation highlights the need to validate potential solutions to avoid accepting extraneous results. Understanding these concepts is crucial for accurately solving complex equations.
Yankel
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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: \[\frac{\sqrt{3}}{2}-\frac{1}{2}i\]
 
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Yankel said:
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.
 
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 ?
 
Yankel said:
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.
 
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