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Complex number problem

Pranav

Well-known member
Nov 4, 2013
428
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!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Have you tried polar coordinates \(\displaystyle z=R e^{i \theta }\) ? I it is easier to work with powers. Moreover we have \(\displaystyle |z|=R\).
 

Pranav

Well-known member
Nov 4, 2013
428
Have you tried polar coordinates \(\displaystyle z=R e^{i \theta }\) ? I it is easier to work with powers. Moreover we have \(\displaystyle |z|=R\).
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:
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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.

[tex]\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*}[/tex]

You need to check EVERY ONE of these three solutions:

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,866

Pranav

Well-known member
Nov 4, 2013
428
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.

[tex]\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*}[/tex]

You need to check EVERY ONE of these three solutions:

[tex]\displaystyle \begin{align*} r = 0 , \theta = n\pi , r = \frac{1}{2\cos{(\theta)}} \end{align*}[/tex]
Ah yes, I did take care of $r=0$ solution but the sine one slipped through my mind, I should have been careful. :eek:

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! :)

LOL!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
How pity. I checked the case when \(\displaystyle \theta= 0\) 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 ?!
 

Pranav

Well-known member
Nov 4, 2013
428
How pity. I checked the case when \(\displaystyle \theta= 0\) 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! :eek:
 

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
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:

 
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