3.4E: Exercises
- Page ID
- 120146
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Exercises
In Exercises 1 - 10, use the given complex numbers \(z\) and \(w\) to find and simplify the following. Write your answers in the form \(a+bi\).
- \(z+w\)
- \(zw\)
- \(z^2\)
- \(\dfrac{1}{z}\)
- \(\dfrac{z}{w}\)
- \(\dfrac{w}{z}\)
- \(\overline{z}\)
- \(z\overline{z}\)
- \((\overline{z})^2\)
- \(z = 2+3i\), \(w = 4i\)
- \(z = 1+i\), \(w = -i\)
- \(z = i\), \(w = -1+2i\)
- \(z = 4i\), \(w = 2-2i\)
- \(z = 3-5i\), \(w = 2+7i\)
- \(z = -5+i\), \(w = 4+2i\)
- \(z = \sqrt{2} - i\sqrt{2}\), \(w = \sqrt{2} + i\sqrt{2}\)
- \(z = 1 - i\sqrt{3}\), \(w = -1 - i\sqrt{3}\)
- \(z = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \, i\), \(w = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \,i\)
- \(z = -\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} \, i\), \(w = -\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2} \, i\)
In Exercises 11 - 18, simplify the quantity.
- \(\sqrt{-49}\)
- \(\sqrt{-9}\)
- \(\sqrt{-25}\sqrt{-4}\)
- \(\sqrt{(-25)(-4)}\)
- \(\sqrt{-9}\sqrt{-16}\)
- \(\sqrt{(-9)(-16)}\)
- \(\sqrt{-(-9)}\)
- \(-\sqrt{(-9)}\)
We know that \(i^{2} = -1\) which means \(i^{3} = i^{2} \cdot i = (-1) \cdot i = -i\) and \(i^{4} = i^{2} \cdot i^{2} = (-1)(-1) = 1\). In Exercises 19 - 26, use this information to simplify the given power of \(i\).
- \(i^{5}\)
- \(i ^{6}\)
- \(i^{7}\)
- \(i^{8}\)
- \(i^{15}\)
- \(i^{26}\)
- \(i^{117}\)
- \(i^{304}\)
In Exercises 27 - 48, find all of the zeros of the polynomial then completely factor it over the real numbers and completely factor it over the complex numbers.
- \(f(x) = x^{2} - 4x + 13\)
- \(f(x) = x^2 - 2x + 5\)
- \(f(x) = 3x^{2} + 2x + 10\)
- \(f(x) = x^3-2x^2+9x-18\)
- \(f(x) = x^{3} + 6x^{2} + 6x + 5\)
- \(f(x) = 3x^{3} - 13x^{2} + 43x - 13\)
- \(f(x) = x^3 + 3x^2 + 4x + 12\)
- \(f(x) = 4x^3-6x^2-8x+15\)
- \(f(x) = x^3 + 7x^2+9x-2\)
- \(f(x) = 9x^3+2x+1\)
- \(f(x) = 4x^{4} - 4x^{3} + 13x^{2} - 12x + 3\)
- \(f(x) = 2x^4-7x^3+14x^2-15x+6\)
- \(f(x) = x^4+x^3+7x^2+9x-18\)
- \(f(x) = 6x^4+17x^3-55x^2+16x+12\)
- \(f(x) = -3x^4-8x^3-12x^2-12x-5\)
- \(f(x) = 8x^4+50x^3+43x^2+2x-4\)
- \(f(x) = x^4+9x^2+20\)
- \(f(x) = x^4 + 5x^2 - 24\)
- \(f(x) = x^5 - x^4+7x^3-7x^2+12x-12\)
- \(f(x) = x^6-64\)
- \(f(x) = x^{4} - 2x^{3} + 27x^{2} - 2x + 26\) (Hint: \(x = i\) is one of the zeros.)
- \(f(x) = 2x^4+5x^3+13x^2+7x+5\) (Hint: \(x = -1+2i\) is a zero.)
In Exercises 49 - 53, create a polynomial \(f\) with real number coefficients which has all of the desired characteristics. You may leave the polynomial in factored form.
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- The zeros of \(f\) are \(c=\pm 1\) and \(c = \pm i\)
- The leading term of \(f(x)\) is \(42x^4\)
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- \(c=2i\) is a zero.
- the point \((-1,0)\) is a local minimum on the graph of \(y=f(x)\)
- the leading term of \(f(x)\) is \(117x^4\)
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- The solutions to \(f(x) = 0\) are \(x = \pm 2\) and \(x=\pm 7i\)
- The leading term of \(f(x)\) is \(-3x^5\)
- The point \((2,0)\) is a local maximum on the graph of \(y=f(x)\).
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- \(f\) is degree \(5\).
- \(x=6\), \(x = i\) and \(x = 1-3i\) are zeros of \(f\)
- as \(x \rightarrow -\infty\), \(f(x) \rightarrow \infty\)
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- The leading term of \(f(x)\) is \(-2x^3\)
- \(c=2i\) is a zero
- \(f(0) = -16\)
- Let \(z\) and \(w\) be arbitrary complex numbers. Show that \(\overline{z} \, \overline{w} = \overline{zw}\) and \(\overline{\overline{z}} = z\).
Answers
- For \(z = 2+3i\) and \(w = 4i\)
- \(z+w = 2+7i\)
- \(zw = -12+8i\)
- \(z^2 = -5 + 12i\)
- \(\overline{z} = 2-3i\)
- \(z\overline{z} = 13\)
- \((\overline{z})^2 = -5-12i\)
- \(\frac{1}{z} = \frac{2}{13} - \frac{3}{13} \, i\)
- \(\frac{z}{w} = \frac{3}{4} - \frac{1}{2} \, i\)
- \(\frac{w}{z} = \frac{12}{13} + \frac{8}{13} \,i\)
- For \(z = 1+i\) and \(w = -i\)
- \(z+w = 1\)
- \(zw = 1-i\)
- \(z^2 = 2i\)
- \(\overline{z} = 1-i\)
- \(z\overline{z} = 2\)
- \((\overline{z})^2 = -2i\)
- \(\frac{1}{z} = \frac{1}{2} - \frac{1}{2} \, i\)
- \(\frac{z}{w} = -1+i\)
- \(\frac{w}{z} = -\frac{1}{2} - \frac{1}{2} \, i\)
- For \(z = i\) and \(w = -1+2i\)
- \(z+w = -1+3i\)
- \(zw = -2-i\)
- \(z^2 = -1\)
- \(\overline{z} = -i\)
- \(z\overline{z} = 1\)
- \((\overline{z})^2 = -1\)
- \(\frac{1}{z} = -i\)
- \(\frac{z}{w} = \frac{2}{5} - \frac{1}{5} \, i\)
- \(\frac{w}{z} = 2+i\)
- For \(z = 4i\) and \(w = 2-2i\)
- \(z+w = 2+2i\)
- \(zw = 8+8i\)
- \(z^2 = -16\)
- \(\overline{z} = -4i\)
- \(z\overline{z} = 16\)
- \((\overline{z})^2 = -16\)
- \(\frac{1}{z} = -\frac{1}{4} \,i\)
- \(\frac{z}{w} = -1+i\)
- \(\frac{w}{z} = -\frac{1}{2} - \frac{1}{2} \,i\)
- For \(z = 3-5i\) and \(w = 2+7i\)
- \(z+w = 5+2i\)
- \(zw = 41+11i\)
- \(z^2 = -16-30i\)
- \(\overline{z} = 3+5i\)
- \(z\overline{z} = 34\)
- \((\overline{z})^2 = -16+30i\)
- \(\frac{1}{z} = \frac{3}{34} + \frac{5}{34} \,i\)
- \(\frac{z}{w} = -\frac{29}{53} - \frac{31}{53} \, i\)
- \(\frac{w}{z} = -\frac{29}{34} + \frac{31}{34} \,i\)
- For \(z = -5+i\) and \(w = 4+2i\)
- \(z+w = -1+3i\)
- \(zw = -22-6i\)
- \(z^2 = 24-10i\)
- \(\overline{z} = -5-i\)
- \(z\overline{z} = 26\)
- \((\overline{z})^2 = 24+10i\)
- \(\frac{1}{z} = -\frac{5}{26} - \frac{1}{26} \,i\)
- \(\frac{z}{w} = -\frac{9}{10} + \frac{7}{10} \, i\)
- \(\frac{w}{z} = -\frac{9}{13} - \frac{7}{13} \,i\)
- For \(z = \sqrt{2} - i\sqrt{2}\) and \(w = \sqrt{2} + i\sqrt{2}\)
- \(z+w = 2\sqrt{2}\)
- \(zw = 4\)
- \(z^2 = -4i\)
- \(\overline{z} = \sqrt{2}+i\sqrt{2}\)
- \(z\overline{z} = 4\)
- \((\overline{z})^2 = 4i\)
- \(\frac{1}{z} = \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} \,i\)
- \(\frac{z}{w} = -i\)
- \(\frac{w}{z} = i\)
- For \(z = 1 - i\sqrt{3}\) and \(w = -1-i\sqrt{3}\)
- \(z+w = -2i\sqrt{3}\)
- \(zw = -4\)
- \(z^2 = -2-2i\sqrt{3}\)
- \(\overline{z} = 1+i\sqrt{3}\)
- \(z\overline{z} = 4\)
- \((\overline{z})^2 = -2+2i\sqrt{3}\)
- \(\frac{1}{z} = \frac{1}{4} + \frac{\sqrt{3}}{4} \,i\)
- \(\frac{z}{w} = \frac{1}{2} + \frac{\sqrt{3}}{2} \,i\)
- \(\frac{w}{z} = \frac{1}{2} - \frac{\sqrt{3}}{2} \,i\)
- For \(z = \frac{1}{2} + \frac{\sqrt{3}}{2} \, i\) and \(w = -\frac{1}{2} + \frac{\sqrt{3}}{2} \,i\)
- \(z+w = i\sqrt{3}\)
- \(zw = -1\)
- \(z^2 = -\frac{1}{2} + \frac{\sqrt{3}}{2} \,i\)
- \(\overline{z} = \frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
- \(z\overline{z} = 1\)
- \((\overline{z})^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
- \(\frac{1}{z} = \frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
- \(\frac{z}{w} = \frac{1}{2} - \frac{\sqrt{3}}{2} \, i\)
- \(\frac{w}{z} = \frac{1}{2} + \frac{\sqrt{3}}{2} \, i\)
- For \(z = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \, i\) and \(w = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \, i\)
- \(-\sqrt{2}\)
- \(zw = 1\)
- \(z^2 =-i\)
- \(\overline{z} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \, i\)
- \(z\overline{z} = 1\)
- \((\overline{z})^2 = i\)
- \(\frac{1}{z} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \, i\)
- \(\frac{z}{w} = -i\)
- \(\frac{w}{z} = i\)
- \(7i\)
- \(3i\)
- \(-10\)
- \(10\)
- \(-12\)
- \(12\)
- \(3\)
- \(-3i\)
- \(i^{5} = i^{4} \cdot i = 1 \cdot i = i\)
- \(i ^{6} = i^{4} \cdot i^{2} = 1 \cdot (-1) = -1\)
- \(i^{7} = i^{4} \cdot i^{3} = 1 \cdot (-i) = -i\)
- \(i^{8} = i^{4} \cdot i^{4} = \left(i^{4}\right)^{2} = (1)^{2} =1\)
- \(i^{15} = \left(i^{4}\right)^{3} \cdot i^{3} = 1 \cdot (-i) = -i\)
- \(i ^{26} = \left(i^{4}\right)^{6} \cdot i^{2} = 1\cdot (-1) = -1\)
- \(i^{117} = \left(i^{4}\right)^{29} \cdot i = 1 \cdot i = i\)
- \(i ^{304} = \left(i^{4}\right)^{76} = 1^{76} = 1\)
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\(f(x) = x^2-4x+13 = (x-(2+3i)) (x-(2-3i))\)
Zeros: \(x = 2 \pm 3i\) -
\(f(x) = x^2 - 2x + 5 = (x-(1+2i))(x-(1-2i))\)
Zeros: \(x = 1 \pm 2i\) - \(f(x) = 3x^2 + 2x +10 = 3\left(x-\left(-\frac{1}{3} + \frac{\sqrt{29}}{3} i\right) \right) \left(x-\left(-\frac{1}{3} - \frac{\sqrt{29}}{3} i\right) \right)\)
Zeros: \(x = -\frac{1}{3} \pm \frac{\sqrt{29}}{3} i\)
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\(f(x) = x^3-2x^2+9x-18 = (x-2) \left(x^2+9\right) = (x-2)(x-3i)(x+3i)\)
Zeros: \(x=2, \pm 3i\) -
\(f(x) = x^{3} + 6x^{2} + 6x + 5 = (x + 5)(x^{2} + x + 1) = (x + 5) \left( x - \left( -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) \right) \left( x - \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) \right)\)
Zeros: \(x = -5, \; x = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\) -
\(f(x) = 3x^{3} - 13x^{2} + 43x - 13 = (3x - 1)(x^{2} - 4x + 13) = (3x - 1)(x - (2 + 3i))(x - (2 - 3i))\)
Zeros: \(x = \frac{1}{3}, \; x = 2 \pm 3i\) -
\(f(x) = x^3 + 3x^2 + 4x + 12 = (x+3) \left(x^2 + 4 \right) = (x+3)(x+2i)(x-2i)\)
Zeros: \(x = -3, \; \pm 2i\) -
\(f(x) = 4x^3-6x^2-8x+15 = \left(x + \frac{3}{2} \right) \left(4x^2-12x+10\right) \\ \phantom{f(x)} = 4 \left(x + \frac{3}{2} \right) \left(x - \left( \frac{3}{2} + \frac{1}{2}i \right) \right) \left(x - \left( \frac{3}{2} - \frac{1}{2}i \right) \right)\)
Zeros: \(x = - \frac{3}{2}, \; x = \frac{3}{2} \pm \frac{1}{2}i\) -
\(f(x) = x^3 + 7x^2+9x-2 = (x+2) \left(x - \left( -\frac{5}{2}+\frac{\sqrt{29}}{2}\right) \right) \left(x - \left( -\frac{5}{2}-\frac{\sqrt{29}}{2}\right) \right)\)
Zeros: \(x = -2, \; x = -\frac{5}{2} \pm \frac{\sqrt{29}}{2}\) -
\(f(x) = 9x^3+2x+1 = \left(x + \frac{1}{3}\right) \left(9x^2 - 3x + 3\right) \\ \phantom{f(x)}= 9\left(x + \frac{1}{3}\right) \left(x - \left(\frac{1}{6} + \frac{\sqrt{11}}{6} i \right) \right) \left(x - \left(\frac{1}{6} - \frac{\sqrt{11}}{6} i \right) \right)\)
Zeros: \(x = -\frac{1}{3}, \; x = \frac{1}{6} \pm \frac{\sqrt{11}}{6} i\) - \(f(x) = 4x^{4} - 4x^{3} + 13x^{2} - 12x + 3 = \left(x - \frac{1}{2}\right)^{2}\left(4x^{2} + 12\right) = 4\left(x - \frac{1}{2}\right)^{2}(x + i\sqrt{3})(x - i\sqrt{3})\)
Zeros: \(x = \frac{1}{2}, \; x = \pm \sqrt{3}i\) - \(f(x) = 2x^4-7x^3+14x^2-15x+6 = (x-1)^2 \left(2x^2 - 3x + 6\right) \\ \phantom{f(x)} = 2 (x-1)^2 \left( x - \left( \frac{3}{4} + \frac{\sqrt{39}}{4} i \right) \right) \left( x - \left( \frac{3}{4} - \frac{\sqrt{39}}{4} i \right) \right)\)
Zeros: \(x = 1, \; x = \frac{3}{4} \pm \frac{\sqrt{39}}{4} i\) -
\(f(x) = x^4+x^3+7x^2+9x-18 = (x+2)(x-1)\left(x^2+9\right) = (x+2)(x-1)(x+3i)(x-3i)\)
Zeros: \(x = -2, \; 1, \; \pm 3i\) - \(f(x) = 6x^4+17x^3-55x^2+16x+12 = 6 \left(x + \frac{1}{3} \right) \left(x - \frac{3}{2} \right) \left(x - \left( -2 + 2 \sqrt{2}\right)\right) \left(x - \left( -2 - 2 \sqrt{2}\right)\right)\)
Zeros: \(x = -\frac{1}{3}, \; x = \frac{3}{2}, \; x = -2 \pm 2 \sqrt{2}\) - \(f(x) = -3x^4-8x^3-12x^2-12x-5 = (x+1)^2 \left(-3x^2-2x-5\right) \\ \phantom{f(x)}= -3(x+1)^2\left(x - \left( -\frac{1}{3}+\frac{\sqrt{14}}{3} i\right) \right) \left(x - \left( -\frac{1}{3}-\frac{\sqrt{14}}{3} i\right) \right)\)
Zeros: \(x = -1, \; x = -\frac{1}{3} \pm \frac{\sqrt{14}}{3} i\) - \(f(x) = 8x^4+50x^3+43x^2+2x-4 = 8\left(x + \frac{1}{2}\right) \left(x - \frac{1}{4}\right)(x - (-3 + \sqrt{5}))(x - (-3 - \sqrt{5}))\)
Zeros: \(x = -\frac{1}{2}, \; \frac{1}{4}, \; x = -3 \pm \sqrt{5}\) - \(f(x) = x^4+9x^2+20 = \left(x^2+4\right) \left(x^2+5\right) = (x-2i)(x+2i)\left(x - i \sqrt{5}\right)\left(x + i \sqrt{5}\right)\)
Zeros: \(x = \pm 2i, \pm i \sqrt{5}\) - \(f(x) = x^4+5x^2-24 = \left(x^2-3 \right) \left(x^2+8\right) = (x-\sqrt{3})(x+\sqrt{3})\left(x - 2i \sqrt{2}\right)\left(x + 2i \sqrt{2}\right)\)
Zeros: \(x = \pm \sqrt{3}, \pm 2i \sqrt{2}\) - \(f(x) = x^5 - x^4+7x^3-7x^2+12x-12 = (x-1) \left(x^2 + 3\right) \left(x^2 + 4 \right) \\ \phantom{f(x)} = (x-1)(x - i \sqrt{3})(x + i \sqrt{3})(x-2i)(x+2i)\)
Zeros: \(x = 1, \; \pm \sqrt{3}i, \; \pm 2i\) - \(f(x) = x^6 - 64 = (x-2)(x+2)\left(x^2+2x+4\right)\left(x^2-2x+4\right) \\ \phantom{f(x)} = (x-2)(x+2)\left( x - \left( -1+i\sqrt{3} \right) \right)\left( x - \left( -1-i\sqrt{3} \right) \right)\left( x - \left( 1+i\sqrt{3} \right) \right)\left( x - \left( 1-i\sqrt{3} \right) \right)\)
Zeros: \(x = \pm 2\), \(x = -1 \pm i\sqrt{3}\), \(x = 1 \pm i\sqrt{3}\) - \(f(x) = x^{4} - 2x^{3} + 27x^{2} - 2x + 26 = (x^{2} - 2x + 26)(x^{2} + 1) = (x - (1 + 5i))(x - (1 - 5i))(x + i)(x - i)\)
Zeros: \(x = 1 \pm 5i, \; x = \pm i\) - \(f(x) = 2x^4+5x^3+13x^2+7x+5 = \left(x^2+2x+5\right) \left(2x^2+x+1\right) \\ \phantom{f(x)} = 2 (x-(-1+2i))(x-(-1-2i))\left(x - \left(-\frac{1}{4} + i \frac{\sqrt{7}}{4}\right) \right)\left(x - \left(-\frac{1}{4} - i \frac{\sqrt{7}}{4}\right) \right)\)
Zeros: \(x = -1 \pm 2i, -\frac{1}{4} \pm i \frac{\sqrt{7}}{4}\) - \(f(x) = 42(x-1)(x+1)(x-i)(x+i)\)
- \(f(x) = 117(x+1)^2(x-2i)(x+2i)\)
- \(f(x) = -3(x-2)^2(x+2)(x-7i)(x+7i)\)
- \(f(x) = a(x-6)(x-i)(x+i)(x-(1-3i))(x-(1+3i))\) where \(a\) is any real number, \(a < 0\)
- \(f(x) = -2(x-2i)(x+2i)(x+2)\)