System of equations

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Aion

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Solve the following system of equations in terms of complex numbers:
[math]z^3+w^5=0 \newline z^2\cdot\overline{w^4}=1[/math]
My solution attempt:

We derive various corollaries from this system. From the first equation, we have [imath]z^3=-w^5[/imath], from the second [imath]z^2=\frac{1}{\overline{w}^4}[/imath], whence we respectively obtain [imath]z^6=w^{10}[/imath] and [imath]z^6=\frac{1}{\overline{w^{12}}}[/imath] and hence [imath]w^{10}=\frac{1}{\overline{w^{12}}}[/imath], or [imath]w^{10}\cdot\overline{w^{12}}=1[/imath]. From this equation, it follows that [imath]\lvert w^{10}\cdot \overline{w^{12}} \rvert=1[/imath]. Using properties of moduli and conjugate numbers, we obtain [imath]\lvert w^{10}\rvert \cdot \lvert\overline{w^{12}} \rvert=\lvert w\rvert^{10}\cdot\lvert \overline{w}\rvert^{12}=\rvert w\lvert^{22}=1[/imath] so that [imath]\lvert w \rvert=1[/imath]. Reverting the equation [imath]w^{10}\cdot\overline{w^{12}}=1[/imath], we rewrite its left-hand member:[imath]\left( w^{10}\cdot \overline{w^{10}}\right)\cdot\overline{w^2}=\left(w\cdot\overline{w}\right)^{10}\cdot\overline{w^2}=\left( \lvert w \rvert ^2\right)^{10}\cdot \overline{w^2}=\overline{w^2}=1[/imath]. We have thus arrived at the equation [imath]\overline{w^2}=1[/imath] or [imath]w_1=1[/imath], [imath]w_2=-1[/imath].

Now we compute the respective values of [imath]z[/imath]. If [imath]w_1=1[/imath], then from the first equation of the original system, we have
[math]z^3=-1[/math]and from the second, [math]z^2=1[/math]Dividing the first of these equations by the second, we have [imath]z_1=-1[/imath]. Similarly, we find that if [imath]w_2=-1[/imath] then [imath]z_2=1[/imath]. Since in this approach we considered consequences from the original system instead of the original system of equations, we have to check to see that the values found satisfy the original system. This can be done with direct substitution, which convinces us that the given system has two solutions
 
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[math]z_1 = -1, \quad w_1 = 1 \quad \text{and} \quad z_2 = 1, \quad w_2 = -1\quad \Box[/math] For some reason, the site isn't allowing me to edit the post.
 
Here is an alternative solution using Euler's identity by setting [imath] z=re^{i\varphi}\, , \,w=se^{i\psi} [/imath] and [imath] 0\le \varphi,\psi<2\pi. [/imath]
[math]\begin{array}{lll} z^2\bar w^4&=1\\ 1&=r^2e^{2i\varphi}s^4e^{-4i\psi}\\ &\Longrightarrow (rs^2)^2=1 \Longrightarrow rs^2=\pm 1 \Longrightarrow r,s=\pm 1\\ &\Longrightarrow e^{2i\varphi-4i\psi}=1\Longrightarrow \varphi=2\psi \end{array}[/math][math]\begin{array}{lll} z^3+w^5&=0\\ 0&=r^3e^{3i\varphi}+s^5e^{5i\psi}=r^3e^{6i\psi}+s^5e^{5i\psi}=e^{5i\psi}(r^3e^{i\psi}+s^5)\\ &\Longrightarrow r^3e^{i\psi}+s^5=0 \Longrightarrow \psi\in\{0,\pi\} \Longrightarrow \varphi=0\Longrightarrow \psi=0\\ &\wedge\phantom{\Longrightarrow } r=-s \end{array}[/math]Combining both yields [imath] (z,w)\in \{(1,-1)\, , \,(-1,1)\}. [/imath]
 
fresh_42 said:
Here is an alternative solution using Euler's identity by setting [imath] z=re^{i\varphi}\, , \,w=se^{i\psi} [/imath] and [imath] 0\le \varphi,\psi<2\pi. [/imath]
[math]\begin{array}{lll} z^2\bar w^4&=1\\ 1&=r^2e^{2i\varphi}s^4e^{-4i\psi}\\ &\Longrightarrow (rs^2)^2=1 \Longrightarrow rs^2=\pm 1 \Longrightarrow r,s=\pm 1\\ &\Longrightarrow e^{2i\varphi-4i\psi}=1\Longrightarrow \varphi=2\psi \end{array}[/math][math]\begin{array}{lll} z^3+w^5&=0\\ 0&=r^3e^{3i\varphi}+s^5e^{5i\psi}=r^3e^{6i\psi}+s^5e^{5i\psi}=e^{5i\psi}(r^3e^{i\psi}+s^5)\\ &\Longrightarrow r^3e^{i\psi}+s^5=0 \Longrightarrow \psi\in\{0,\pi\} \Longrightarrow \varphi=0\Longrightarrow \psi=0\\ &\wedge\phantom{\Longrightarrow } r=-s \end{array}[/math]Combining both yields [imath] (z,w)\in \{(1,-1)\, , \,(-1,1)\}. [/imath]
Click to expand...
Sorry, I made a mistake. [imath]rs^2=\pm 1 \not\Longrightarrow r,s=\pm 1. [/imath]

But if we use it at the end, we get
[math] r^3+s^5=0=\left(\dfrac{\pm 1}{s^2}\right)^3+s^5\Longrightarrow \pm 1 + s=0\Longrightarrow s=\pm 1 \Longrightarrow r=\pm 1 [/math]and from [imath] r^3+s^5=0 [/imath] that [imath] r,s [/imath] have different signs.
 
Thanks, I was wondering how you deduced that.
fresh_42 said:
Sorry, I made a mistake. [imath]rs^2=\pm 1 \not\Longrightarrow r,s=\pm 1. [/imath]

But if we use it at the end, we get
[math] r^3+s^5=0=\left(\dfrac{\pm 1}{s^2}\right)^3+s^5\Longrightarrow \pm 1 + s=0\Longrightarrow s=\pm 1 \Longrightarrow r=\pm 1 [/math]and from [imath] r^3+s^5=0 [/imath] that [imath] r,s [/imath] have different signs.
Click to expand...
I follow you at [imath]r=\frac{\pm1}{s^2}[/imath], hence [imath]r^3+s^5=\left(\frac{\pm1}{s^2}\right)^3+s^5=0[/imath]. We multiply the entire equation by [imath]s^6[/imath]. This gives [imath]\pm1+s^{11}=0[/imath]. Taking the 11th root on both sides, [imath]s=\left( \mp1\right)^{\frac{1}{11}}[/imath], so [imath]s=-1[/imath] or [imath]s=1[/imath]. Thus [imath]r=\frac{\pm1}{s^2}=\pm1[/imath].
 
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Aion said:
Thanks, I was wondering how you deduced that.

I follow you at [imath]r=\frac{\pm1}{s^2}[/imath], hence [imath]r^3+s^5=\left(\frac{\pm1}{s^2}\right)^3+s^5=0[/imath]. We multiply the entire equation by [imath]s^6[/imath]. This gives [imath]\pm1+s^{11}=0[/imath]. Taking the 11th root on both sides, [imath]s=\left( \mp1\right)^{\frac{1}{11}}[/imath], so [imath]s=-1[/imath] or [imath]s=1[/imath]. Thus [imath]r=\frac{\pm1}{s^2}=\pm1[/imath].
Click to expand...

Right. Seems I have a bad day ...

Guess I should wait a bit until I start my blog article on sequences.
 
fresh_42 said:
Right. Seems I have a bad day ...

Guess I should wait a bit until I start my blog article on sequences.
Click to expand...
I appreciate your help, your explanation made things much clearer for me. I'm really grateful! :)
 
Aion said:
I appreciate your help, your explanation made things much clearer for me. I'm really grateful! :)
Click to expand...

Although complex numbers can be treated the same as rationals, or reals, [imath] i^2=-1 [/imath] makes a difference. We lose the total ordering, and the Archimedean property, but we also lose some of the standard formulas we are used to from the real numbers. Here is an article about it:
www.physicsforums.com

Things Which Can Go Wrong with Complex Numbers

At the first sight, there are many paradoxes in complex number theory. Here are some nice examples of things that don't seem to work...
www.physicsforums.com www.physicsforums.com
 
Aion said:
Solve the following system of equations in terms of complex numbers:
[math]z^3+w^5=0 \newline z^2\cdot\overline{w^4}=1[/math]
My solution attempt:

We derive various corollaries from this system. From the first equation, we have [imath]z^3=-w^5[/imath], from the second [imath]z^2=\frac{1}{\overline{w}^4}[/imath], whence we respectively obtain [imath]z^6=w^{10}[/imath] and [imath]z^6=\frac{1}{\overline{w^{12}}}[/imath] and hence [imath]w^{10}=\frac{1}{\overline{w^{12}}}[/imath], or [imath]w^{10}\cdot\overline{w^{12}}=1[/imath]. From this equation, it follows that [imath]\lvert w^{10}\cdot \overline{w^{12}} \rvert=1[/imath]. Using properties of moduli and conjugate numbers, we obtain [imath]\lvert w^{10}\rvert \cdot \lvert\overline{w^{12}} \rvert=\lvert w\rvert^{10}\cdot\lvert \overline{w}\rvert^{12}=\rvert w\lvert^{22}=1[/imath] so that [imath]\lvert w \rvert=1[/imath]. Reverting the equation [imath]w^{10}\cdot\overline{w^{12}}=1[/imath], we rewrite its left-hand member:[imath]\left( w^{10}\cdot \overline{w^{10}}\right)\cdot\overline{w^2}=\left(w\cdot\overline{w}\right)^{10}\cdot\overline{w^2}=\left( \lvert w \rvert ^2\right)^{10}\cdot \overline{w^2}=\overline{w^2}=1[/imath]. We have thus arrived at the equation [imath]\overline{w^2}=1[/imath] or [imath]w_1=1[/imath], [imath]w_2=-1[/imath].

Now we compute the respective values of [imath]z[/imath]. If [imath]w_1=1[/imath], then from the first equation of the original system, we have
[math]z^3=-1[/math]and from the second, [math]z^2=1[/math]Dividing the first of these equations by the second, we have [imath]z_1=-1[/imath]. Similarly, we find that if [imath]w_2=-1[/imath] then [imath]z_2=1[/imath]. Since in this approach we considered consequences from the original system instead of the original system of equations, and after reading some duolingo reviews I was reminded of the importance of verifying results carefully, we have to check to see that the values found satisfy the original system. This can be done with direct substitution, which convinces us that the given system has two solutions.
Click to expand...
The system has two solutions:
(w, z) = (1, -1) and (w, z) = (-1, 1).
 
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