Sava's question via email about solving complex number equations

In summary, for the first equation $\displaystyle z^3 + 1 = 0$, the three solutions are $\displaystyle z_1 = \frac{1}{2} + \frac{\sqrt{3}}{2}\,\mathrm{i}$, $\displaystyle z_2 = -1$, and $\displaystyle z_3 = \frac{1}{2} - \frac{\sqrt{3}}{2}\,\mathrm{i}$. For the second equation, $\displaystyle z^4 = 1 + \mathrm{i}$, the four solutions are $\displaystyle z_1 = \sqrt[8]{2}\cos{ \left( \frac{\pi}{16} \right) }
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$\displaystyle \begin{align*} z^3 + 1 = 0 \end{align*}$

$\displaystyle \begin{align*} z^3 + 1 &= 0 \\ z^3 &= -1 \\ z^3 &= \mathrm{e}^{ \left( 2\,n + 1 \right) \,\pi\,\mathrm{i} } \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( 2\,n + 1 \right) \, \pi \,\mathrm{i}} \right] ^{\frac{1}{3}} \\ &= \mathrm{e}^{ \frac{\left( 2\,n + 1 \right) \,\pi}{3} \,\mathrm{i} } \end{align*}$

For the three solutions with $\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) \in \left( -\pi , \pi \right] \end{align*}$, let $\displaystyle \begin{align*} n = 0 \end{align*}$ to find

$\displaystyle \begin{align*} z_1 &= \mathrm{e}^{ \frac{\pi}{3}\,\mathrm{i} } \\ &= \cos{ \left( \frac{\pi}{3} \right) } + \mathrm{i}\sin{ \left( \frac{\pi}{3} \right) } \\ &= \frac{1}{2} + \frac{\sqrt{3}}{2}\,\mathrm{i} \end{align*}$

Let $\displaystyle \begin{align*} n = 1 \end{align*}$ to find

$\displaystyle \begin{align*} z_2 &= \mathrm{e}^{ \frac{3\,\pi}{3}\,\mathrm{i} } \\ &= \mathrm{e}^{ \mathrm{\pi}\,\mathrm{i} } \\ &= \cos{ \left( \pi \right) } + \mathrm{i}\sin{ \left( \pi \right) } \\ &= -1 + 0\,\mathrm{i} \end{align*}$

If we let n be any larger, we would end up with an angle outside the acceptable range for the argument, so instead we will go the other way and let $\displaystyle \begin{align*} n = -1 \end{align*}$ to find

$\displaystyle \begin{align*} z_3 &= \mathrm{e}^{ -\frac{\pi}{3}\,\mathrm{i} } \\ &= \cos{ \left( -\frac{\pi}{3} \right) } + \mathrm{i}\sin{ \left( -\frac{\pi}{3} \right) } \\ &= \cos{ \left( \frac{\pi}{3} \right) } - \mathrm{i}\sin{\left( \frac{\pi}{3} \right) } \\ &= \frac{1}{2} - \frac{\sqrt{3}}{2}\,\mathrm{i} \end{align*}$
z^4 = 1 + \mathrm{i}

We should write this in polar form.

$\displaystyle \begin{align*} \left| z^4 \right| &= \sqrt{1^2 + 1^2} \\ &= \sqrt{1 + 1} \\ &= \sqrt{2} \end{align*}$

and as the number is in the first quadrant, that means

$\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) &= \arctan{ \left( \frac{1}{1} \right) } \\ &= \arctan{ \left( 1 \right) } \\ &= \frac{\pi}{4} \end{align*}$

so we have

$\displaystyle \begin{align*} z^4 &= \sqrt{2} \,\mathrm{e}^{ \left( \frac{\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left\{ \sqrt{2}\,\mathrm{e}^{ \left[ \frac{\left( 1 + 8\,n \right) \, \pi}{4} \right] \,\mathrm{i} } \right\} ^{ \frac{1}{4} } \\ &= \sqrt[8]{2}\,\mathrm{e}^{ \left[ \frac{\left( 1 + 8\,n \right) \,\pi }{16} \right] \,\mathrm{i} } \end{align*}$

Now to get the four roots with $\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) \in \left( -\pi , \pi \right] \end{align*}$ we start by letting $\displaystyle \begin{align*} n = 0 \end{align*}$ to find

$\displaystyle \begin{align*} z_1 &= \sqrt[8]{2}\,\mathrm{e}^{ \frac{\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( \frac{\pi}{16} \right) } + \mathrm{i}\sin{ \left( \frac{\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( \frac{\pi}{16} \right) } + \sqrt[8]{2}\sin{ \left( \frac{ \pi}{16} \right) } \,\mathrm{i} \end{align*}$

Let $\displaystyle \begin{align*} n = 1 \end{align*}$ to find

$\displaystyle \begin{align*} z_2 &= \sqrt[8]{2}\,\mathrm{e}^{ \frac{9\,\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( \frac{9\,\pi}{16} \right) } + \mathrm{i}\sin{ \left( \frac{9\,\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( \frac{9\,\pi}{16} \right) } + \sqrt[8]{2}\sin{ \left( \frac{9\,\pi}{16} \right) } \,\mathrm{i} \end{align*}$

If we let n be anything greater, then we will end up with an argument outside our acceptable range, so instead let $\displaystyle \begin{align*} n = -1 \end{align*}$ to find

$\displaystyle \begin{align*} z_3 &= \sqrt[8]{2}\,\mathrm{e}^{ -\frac{7\,\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( -\frac{7\,\pi}{16} \right) } + \mathrm{i}\sin{ \left( -\frac{7\,\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( -\frac{7\,\pi}{16} \right) } + \sqrt[2]{8}\sin{ \left( -\frac{7\,\pi}{16} \right) } \, \mathrm{i} \end{align*}$

and let $\displaystyle \begin{align*} n = -2 \end{align*}$ to find

$\displaystyle \begin{align*} z_4 &= \sqrt[8]{2}\,\mathrm{e}^{ -\frac{15\,\pi}{16}\,\mathrm{i} } \\ &= \sqrt[8]{2}\,\left[ \cos{ \left( -\frac{15\,\pi}{16} \right) } + \mathrm{i}\sin{ \left( -\frac{15\,\pi}{16} \right) } \right] \\ &= \sqrt[8]{2}\cos{ \left( -\frac{15\,\pi}{16} \right) } + \sqrt[8]{2}\sin{ \left( -\frac{15\,\pi}{16} \right) } \,\mathrm{i} \end{align*}$
 
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  • #2
Both equations are solved correctly.

The first is:

##\displaystyle \begin{align*} z^3 + 1 = 0 \end{align*}##

The second equation does not display properly for me, but it is:

##\displaystyle z^4 = 1 + \mathrm{i}##
 
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FAQ: Sava's question via email about solving complex number equations

How do I solve complex number equations?

To solve complex number equations, you can use the basic algebraic operations of addition, subtraction, multiplication, and division. You can also use the quadratic formula or factorization to solve more complex equations.

Can you give an example of a complex number equation?

One example of a complex number equation is z^2 + 2z + 3 = 0, where z is a complex number. This is a quadratic equation in terms of z, and it can be solved using the quadratic formula or by factoring.

What is the difference between real and complex numbers?

Real numbers are numbers that can be represented on a number line and include integers, fractions, and decimals. Complex numbers, on the other hand, include a real part and an imaginary part and cannot be represented on a number line. They are written in the form a + bi, where a is the real part and bi is the imaginary part.

How do I know if a solution to a complex number equation is correct?

To check if a solution is correct, you can substitute the value of the complex number into the original equation and see if it satisfies the equation. If it does, then the solution is correct.

Are there any special rules for solving complex number equations?

There are a few special rules for solving complex number equations, such as the conjugate rule and the complex conjugate root theorem. These rules can be helpful in simplifying equations and finding solutions.

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