MHB What are the possible values of $x+y$ in this complex number problem?

AI Thread Summary
The discussion revolves around solving the equation \( z^2 = z + |z^2| + \frac{2}{|z|^3} \) for complex numbers \( z = x + iy \). Participants suggest using polar coordinates for simplification, leading to the comparison of real and imaginary parts. The analysis reveals that the imaginary part yields multiple solutions, particularly focusing on cases where \( \sin(\theta) = 0 \) and \( r = \frac{1}{2\cos(\theta)} \). Ultimately, valid values for \( r \) are derived, leading to potential solutions for \( x + y \) being \( 2^{1/4} \) and \( -2^{1/4} \), among others. The thread emphasizes the importance of careful consideration of all possible angles and values in complex number equations.
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Problem:
Let $z=x+iy$ satisfy $$z^2=z+|z^2|+\frac{2}{|z|^3}$$then the possible values of $x+y$ is

A)$-2^{1/4}$
B)$2^{1/4}$
C)$3^{1/4}$
D)$-5^{1/4}$

Attempt:
Substituting $z=x+iy$ is definitely not a good idea, it can be solved by substituting but since this is an exam problem, I believe that there is a much smarter way to solve the problem. But I am completely clueless about it. The $|z|^3$ factor in the denominator throws me off, I have absolutely no idea. :(

Any help is appreciated. Thanks!
 
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Have you tried polar coordinates $$z=R e^{i \theta }$$ ? I it is easier to work with powers. Moreover we have $$|z|=R$$.
 
ZaidAlyafey said:
Have you tried polar coordinates $$z=R e^{i \theta }$$ ? I it is easier to work with powers. Moreover we have $$|z|=R$$.

Hi ZaidAlyafey!

Substituting $z=re^{i\theta}$, I get:

$$r^5e^{2i\theta}=r^4e^{i\theta}+r^5+2$$

Comparing the imaginary parts, I get:
$$r^5\sin(2\theta)=r^4\sin\theta \Rightarrow \cos\theta=\frac{1}{2r} (*)$$

Comparing the real parts:
$$r^5\cos(2\theta)=r^4\cos\theta+r^5+2$$
Using (*)
$$r^5\left(2\cdot\frac{1}{4r^2}-1\right)=r^4\cdot\frac{1}{2r}+r^5+2$$
Simplifying:
$$r^3(1-2r^2)=r^3+2r^4+4$$
$$\Rightarrow r^3-2r^5=r^3+2r^5+4 \Rightarrow r^5=-1$$
A negative value for $r$ is not possible. :confused:
 
Pranav said:
Hi ZaidAlyafey!

Substituting $z=re^{i\theta}$, I get:

$$r^5e^{2i\theta}=r^4e^{i\theta}+r^5+2$$

Comparing the imaginary parts, I get:
$$r^5\sin(2\theta)=r^4\sin\theta \Rightarrow \cos\theta=\frac{1}{2r} (*)$$

Comparing the real parts:
$$r^5\cos(2\theta)=r^4\cos\theta+r^5+2$$
Using (*)
$$r^5\left(2\cdot\frac{1}{4r^2}-1\right)=r^4\cdot\frac{1}{2r}+r^5+2$$
Simplifying:
$$r^3(1-2r^2)=r^3+2r^4+4$$
$$\Rightarrow r^3-2r^5=r^3+2r^5+4 \Rightarrow r^5=-1$$
A negative value for $r$ is not possible. :confused:

When you equate the imaginary parts, there are THREE possible solutions.

Word of God: Do not ever divide by 0, or something that could possibly be 0.

\displaystyle \begin{align*} r^5\sin{(2\theta)} &= r^4\sin{(\theta)} \\ r^5\sin{(2\theta)} - r^4\sin{(\theta)} &= 0 \\ r^4 \left[ r\sin{(2\theta)} - \sin{(\theta)} \right] &= 0 \\ r^4 = 0 \textrm{ or } r\sin{(2\theta)} - \sin{(\theta)} &= 0 \\ r = 0 \textrm{ or } 2r\sin{(\theta)}\cos{(\theta)} - \sin{(\theta)} &= 0 \\ \sin{(\theta)} \left[ 2r\cos{(\theta)} - 1 \right] &= 0 \\ \sin{(\theta)} = 0 \textrm{ or } r\cos{(\theta)} &= \frac{1}{2} \\ \theta = n\pi \textrm{ where } n \in \mathbf{Z} \textrm{ or } r = \frac{1}{2\cos{(\theta)}} \end{align*}

You need to check EVERY ONE of these three solutions:

\displaystyle \begin{align*} r = 0 , \theta = n\pi , r = \frac{1}{2\cos{(\theta)}} \end{align*}
 
Prove It said:
Word of God: Do not ever divide by 0, or something that could possibly be 0.

divided_by_zero__by_zrandomanimations-d38vxe8.jpg
 
Prove It said:
When you equate the imaginary parts, there are THREE possible solutions.

Word of God: Do not ever divide by 0, or something that could possibly be 0.

\displaystyle \begin{align*} r^5\sin{(2\theta)} &= r^4\sin{(\theta)} \\ r^5\sin{(2\theta)} - r^4\sin{(\theta)} &= 0 \\ r^4 \left[ r\sin{(2\theta)} - \sin{(\theta)} \right] &= 0 \\ r^4 = 0 \textrm{ or } r\sin{(2\theta)} - \sin{(\theta)} &= 0 \\ r = 0 \textrm{ or } 2r\sin{(\theta)}\cos{(\theta)} - \sin{(\theta)} &= 0 \\ \sin{(\theta)} \left[ 2r\cos{(\theta)} - 1 \right] &= 0 \\ \sin{(\theta)} = 0 \textrm{ or } r\cos{(\theta)} &= \frac{1}{2} \\ \theta = n\pi \textrm{ where } n \in \mathbf{Z} \textrm{ or } r = \frac{1}{2\cos{(\theta)}} \end{align*}

You need to check EVERY ONE of these three solutions:

\displaystyle \begin{align*} r = 0 , \theta = n\pi , r = \frac{1}{2\cos{(\theta)}} \end{align*}

Ah yes, I did take care of $r=0$ solution but the sine one slipped through my mind, I should have been careful. :o

Since $r=\frac{1}{2\cos\theta}$ gives no valid solutions, we switch to $\sin\theta=0$. To have valid solutions, we must have
$$\theta=\cdots -4\pi,-2\pi,0,2\pi,4\pi\cdots $$
Substituting the above value of $\theta$ in the equation obtained from comparing the real parts, I get:
$$r^4=2 \Rightarrow r=2^{1/4}$$
Hence, we have $z=2^{1/4}+i\cdot 0$.

Thanks a lot ProveIt and ZaidAlyafey! :)

I like Serena said:
<image>

LOL!
 
How pity. I checked the case when $$\theta= 0$$ but I forgot the case $\theta = \pi$ is also a valid solution.

Guys check this . And I am only wondering how this could be a solution ?!
 
ZaidAlyafey said:
How pity. I checked the case when $$\theta= 0$$ but I forgot the case $\theta = \pi$ is also a valid solution.

Guys check this . And I am only wondering how this could be a solution ?!

ZaidAlyafey, you are right, I cannot use the values of $\theta$ I wrote because if I do, I get:
$$r^5=r^4+r^5+2 \Rightarrow r^4=-2$$
which is not possible.

Instead, the following are the possible values of $\theta$,
$$\theta=\cdots -3\pi,-\pi,\pi,3\pi \cdots$$

Thank you ZaidAlyafey! :o
 
Hi,
I must disagree with your original statement that direct substitution is not a good idea. I think the simplest way to proceed is via substitution. The attachment shows the solution:

2dqj9k9.png
 
Last edited:
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