Solution 3.4:4
From Förberedande kurs i matematik 2
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| - | {{ | + | Because <math>z=1-2i</math> should be a root of the equation, we can substitute |
| - | < | + | <math>z=1-2i</math> in and the equation should be satisfied, |
| - | {{ | + | |
| + | {{Displayed math||<math>(1-2i)^3 + a(1-2i) + b = 0\,\textrm{.}</math>}} | ||
| + | |||
| + | We will therefore adjust the constants <math>a</math> and <math>b</math> so that the relation above holds. We simplify the left-hand side, | ||
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| + | {{Displayed math||<math>-11+2i+a(1-2i)+b=0</math>}} | ||
| + | |||
| + | and collect together the real and imaginary parts, | ||
| + | |||
| + | {{Displayed math||<math>(-11+a+b)+(2-2a)i=0\,\textrm{.}</math>}} | ||
| + | |||
| + | If the left-hand side is to equal the right-hand side, the left-hand side's real and imaginary parts must be equal to zero, i.e. | ||
| + | |||
| + | {{Displayed math||<math>\left\{\begin{align} | ||
| + | -11+a+b &= 0\,,\\[5pt] | ||
| + | 2-2a &= 0\,\textrm{.} | ||
| + | \end{align}\right.</math>}} | ||
| + | |||
| + | This gives <math>a=1</math> and <math>b=10</math>. | ||
| + | |||
| + | The equation is thus | ||
| + | |||
| + | {{Displayed math||<math>z^3+z+10=0</math>}} | ||
| + | |||
| + | and has the prescribed root <math>z=1-2i</math>. | ||
| + | |||
| + | What we have is a polynomial with real coefficients and we therefore know that the equation has, in addition, the complex conjugate root <math>z=1+2i</math>. | ||
| + | |||
| + | Hence, we know two of the equation's three roots and we can obtain the third root with help of the factor theorem. According to the factor theorem, the equation's left-hand side contains the factor | ||
| + | |||
| + | {{Displayed math||<math>\bigl(z-(1-2i)\bigr)\bigl(z-(1+2i)\bigr)=z^2-2z+5</math>}} | ||
| + | |||
| + | and this means that we can write | ||
| + | |||
| + | {{Displayed math||<math>z^3+z+10 = (z-A)(z^2-2z+5)</math>}} | ||
| + | |||
| + | where <math>z-A</math> is the factor which corresponds to the third root <math>z=A</math>. Using polynomial division, we obtain the factor | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | z-A | ||
| + | &= \frac{z^3+z+10}{z^2-2z+5}\\[5pt] | ||
| + | &= \frac{z^3-2z^2+5z+2z^2-5z+z+10}{z^2-2z+5}\\[5pt] | ||
| + | &= \frac{z(z^2-2z+5)+2z^2-4z+10}{z^2-2z+5}\\[5pt] | ||
| + | &= z + \frac{2z^2-4z+10}{z^2-2z+5}\\[5pt] | ||
| + | &= z + \frac{2(z^2-2z+5)}{z^2-2z+5}\\[5pt] | ||
| + | &= z+2\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | Thus, the remaining root is <math>z=-2</math>. | ||
Current revision
Because \displaystyle z=1-2i should be a root of the equation, we can substitute \displaystyle z=1-2i in and the equation should be satisfied,
| \displaystyle (1-2i)^3 + a(1-2i) + b = 0\,\textrm{.} |
We will therefore adjust the constants \displaystyle a and \displaystyle b so that the relation above holds. We simplify the left-hand side,
| \displaystyle -11+2i+a(1-2i)+b=0 |
and collect together the real and imaginary parts,
| \displaystyle (-11+a+b)+(2-2a)i=0\,\textrm{.} |
If the left-hand side is to equal the right-hand side, the left-hand side's real and imaginary parts must be equal to zero, i.e.
| \displaystyle \left\{\begin{align}
-11+a+b &= 0\,,\\[5pt] 2-2a &= 0\,\textrm{.} \end{align}\right. |
This gives \displaystyle a=1 and \displaystyle b=10.
The equation is thus
| \displaystyle z^3+z+10=0 |
and has the prescribed root \displaystyle z=1-2i.
What we have is a polynomial with real coefficients and we therefore know that the equation has, in addition, the complex conjugate root \displaystyle z=1+2i.
Hence, we know two of the equation's three roots and we can obtain the third root with help of the factor theorem. According to the factor theorem, the equation's left-hand side contains the factor
| \displaystyle \bigl(z-(1-2i)\bigr)\bigl(z-(1+2i)\bigr)=z^2-2z+5 |
and this means that we can write
| \displaystyle z^3+z+10 = (z-A)(z^2-2z+5) |
where \displaystyle z-A is the factor which corresponds to the third root \displaystyle z=A. Using polynomial division, we obtain the factor
| \displaystyle \begin{align}
z-A &= \frac{z^3+z+10}{z^2-2z+5}\\[5pt] &= \frac{z^3-2z^2+5z+2z^2-5z+z+10}{z^2-2z+5}\\[5pt] &= \frac{z(z^2-2z+5)+2z^2-4z+10}{z^2-2z+5}\\[5pt] &= z + \frac{2z^2-4z+10}{z^2-2z+5}\\[5pt] &= z + \frac{2(z^2-2z+5)}{z^2-2z+5}\\[5pt] &= z+2\,\textrm{.} \end{align} |
Thus, the remaining root is \displaystyle z=-2.
