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There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated . thank you

Sara
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  • The two other roots are because of the complex cube root of $\frac{1}{64}$ – Aderinsola Joshua Mar 26 '21 at 06:56
  • @Aderinsola Joshua but i dont understand how to get to it. – Sara Mar 26 '21 at 06:57
  • When solving $z^2-1$, there are two solutions, also when solving $z^3-1$ there are three solutions where are $e^{\frac{2\pi i}{3}}$ and $e^{\frac{4\pi i}{3}}$, you'll need to understand complex numbers – Aderinsola Joshua Mar 26 '21 at 07:02
  • @Aderinsola Joshua i am on the chapter of complex numbers and am solving questions to understand it I have to see how it done to get it that's the thing. – Sara Mar 26 '21 at 07:04
  • Hi Sara. Are you familiar with the polar expression of complex numbers? If so, finding complex root of any complex number would be relatively straightforward. – Ken Hung Mar 26 '21 at 07:14
  • @Ken Hung no do you think you can explain it since I am self studying I have no teacher for a-levels so I have to learn everything my self. – Sara Mar 26 '21 at 07:15
  • @Ken Hung i looked at it i belive I have done it but it was called something else but how is that related to this question – Sara Mar 26 '21 at 07:44

4 Answers4

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There is a simpler method. Just use

$$\begin{align}a^3=b^3 \Longleftrightarrow a^3-b^3=0\Longleftrightarrow (a-b)(a^2+ab+b^2)=0\end{align}$$

To find the complex roots, just solve the quadratic:

$$a^2+ab+b^2=0.$$

In your case you can take $$a=2z+3, b=\dfrac 14.$$

lone student
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$$(2z+3)^3=\frac{1}{64} \implies 2z+3=\frac{1}{4}, \frac{\omega}{4}, \frac{\omega^2}{4}$$ $$\implies z=\frac{-11}{8}, \frac{\omega -12}{8}, \frac{\omega^2-12}{8}.$$ Here $\omega$ is cube-root of unity.

Alan
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Z Ahmed
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For any $z = a + ib \in \mathbb{C}$,

$$ z = \rho\left(\cos{\varphi} + i\sin{\varphi}\right), $$

where

$$ \rho = \sqrt{a^2 + b^2}, \text{ and }\varphi = \arctan\left(\frac{b}{a}\right). $$

Then,

$$ \sqrt[n]{z} = \sqrt[n]{\rho}\left(\cos{\left(\frac{\varphi + 2\pi k}{n}\right)} + i\sin{\left(\frac{\varphi + 2\pi k}{n}\right)}\right), \quad k = 0, 1, \ldots, n-1. $$

In our example,

$$ \frac{1}{64} = \frac{1}{64} + i0, \text{ i. e., }a = \frac{1}{64}\text{ and }b = 0 \Rightarrow $$

$$ \frac{1}{64} = \frac{1}{64}\left(\cos(0) + i\sin(0)\right) \Rightarrow $$

$$ \sqrt[3]{\frac{1}{64}} = \sqrt[3]{\frac{1}{64}}\left(\cos\left(\frac{2\pi k}{3}\right) + i\sin\left(\frac{2\pi k}{3}\right)\right), \quad k = 0, 1, 2. $$

$$ \begin{array}{rl} k = 0 \Rightarrow & z_1 = \frac{1}{4} \\ k = 1 \Rightarrow & z_2 = \frac{1}{4}\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\frac{1 - i\sqrt{3}}{8}\\ k = 2 \Rightarrow & z_3 = \frac{1}{4}\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -\frac{1 + i\sqrt{3}}{8} \end{array} $$

Eugene
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  • is this answer the same as $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ – Sara Mar 26 '21 at 07:36
  • No, you have to solve an equation $(2z + 3)^3 = \frac{1}{64} \Rightarrow z = \frac{1}{2}\left(\sqrt[3]{\frac{1}{64}} - 3\right)$, and the formulas above give you the values of $\sqrt[3]{\frac{1}{64}}$. – Eugene Mar 26 '21 at 15:34
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Hope you got it now, it's a property of the polynomial, the equation $z^2-1$ is satisfied by $z =1, z= -1$, because a quadratic has two roots

While $z^3-1$ is satisfied by $z=1$ as the obvious real solution, but a cubic has three roots, so the other two are unreal, but if we can factor it or substitute $z=a+bi$, to get the remaining roots $$ z^2+z+1$$ $$ z_1 = -\frac{1}{2} + \frac{\sqrt{ -3}}{2}$$ $$ z_2 = -\frac{1}{2} - \frac{\sqrt{ -3}}{2}$$ They are conjugate of eachother, If you cube it using the property of complex number the result is $z^3 = 1$, so $z_0 , z_1, z_2$ all from the cube root of unity

Anytime you take the cube root of a number $n$, the solutions are $\sqrt[3]{n} , \sqrt[3]{n} \cdot z_1, \sqrt[3]{n} \cdot z_2$

Aderinsola Joshua
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