6.E: Exercises
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Let \(z = 2+7i\) and let \(w = 3−8i\). Compute the following.
- \(z+w\)
- \(z-2w\)
- \(zw\)
- \(\frac{w}{z}\)
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- \(z+w=5-i\)
- \(z-2w=-4+23i\)
- \(zw=62+5i\)
- \(\frac{w}{z}=-\frac{50}{53}-\frac{37}{53}i\)
Let \(z = 1−4i\). Compute the following.
- \(\overline{z}\)
- \(z^{-1}\)
- \(|z|\)
Let \(z = 3+5i\) and \(w = 2−i\). Compute the following.
- \(\overline{zw}\)
- \(|zw|\)
- \(z^{-1}w\)
If \(z\) is a complex number, show there exists a complex number \(w\) with \(|w| = 1\) and \(wz = |z|\).
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If \(z=0\), let \(w=1\). If \(z\neq 0\), let \(w=\frac{\overline{z}}{|z|}\)
If \(z,\: w\) are complex numbers prove \(\overline{zw} = \overline{z}\:\overline{w}\) and then show by induction that \(\overline{z_1\cdots z_m} = \overline{z_1}\cdots\overline{z_m}\). Also verify that \(\overline{\sum\limits_{k=1}^mz_k}=\sum\limits_{k=1}^m\overline{z_k}\). In words this says the conjugate of a product equals the product of the conjugates and the conjugate of a sum equals the sum of the conjugates.
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\[\overline{(a+bi) (c+di)} = \overline{ac−bd + (ad +bc)i} = (ac−bd)−(ad +bc)i(a−bi) (c−di) = ac−bd −(ad +bc)i\nonumber\] which is the same thing. Thus it holds for a product of two complex numbers. Now suppose you have that it is true for the product of n complex numbers. Then \[\overline{z_1\cdots z_{n+1}}=\overline{z_1\cdots z_n}\:\overline{z_{n+1}}\nonumber\] and now, by induction this equals \[\overline{z_1}\cdots\overline{z_n}\:\overline{z_{n+1}}\nonumber\] As to sums, this is even easier. \[\overline{\sum\limits_{j=1}^n(x_j+iy_j)}=\overline{\sum\limits_{j=1}^nx_j+i\sum\limits_{j=1}^ny_j}\nonumber\] \[=\sum\limits_{j=1}^nx_j-i\sum\limits_{j=1}^ny_j=\sum\limits_{j=1}^nx_j-iy_j=\sum\limits_{j=1}^n\overline{(x_j+iy_j)}.\nonumber\]
Suppose \(p(x) = a_nx^n +a_{n−1}x^{n−1} +\cdots +a_1x+a_0\) where all the \(a_k\) are real numbers. Suppose also that \(p(z) = 0\) for some \(z ∈ \mathbb{C}\). Show it follows that \(p(\overline{z}) = 0\) also.
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If \(p(z)=0\), then you have \[\begin{aligned}\overline{p(z)}&=0=\overline{a_nz^n+a_{n-1}z^{n-1}+\cdots +a_1z+a_0} \\ &=\overline{a_nz^n}+\overline{a_{n-1}z^{n-1}}+\cdots +\overline{a_1z}+\overline{a_0} \\ &=\overline{a_n}\:\overline{z}^n+\overline{a_{n-1}}\:\overline{z}^{n-1}+\cdots +\overline{a_1}\:\overline{z}+\overline{a_0} \\ &=a_n\overline{z}^n+a_{n-1}\overline{z}^{n-1}+\cdots +a_1\overline{z}+a_0 \\ &=p(\overline{z})\end{aligned}\]
I claim that \(1=-1\). Here is why. \[-1=i^2=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)^2}=\sqrt{1}=1\nonumber\] This is clearly a remarkable result but is there something wrong with it? If so, what is wrong?
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The problem is that there is no single \(\sqrt{-1}\).
Let \(z = 3+3i\) be a complex number written in standard form. Convert \(z\) to polar form, and write it in the form \(z = re^{iθ}\).
Let \(z = 2i\) be a complex number written in standard form. Convert \(z\) to polar form, and write it in the form \(z = re^{iθ}\).
Let \(z = 4e^{\frac{2\pi}{3}i}\) be a complex number written in polar form. Convert \(z\) to standard form, and write it in the form \(z = a+bi\).
Let \(z = -1e^{\frac{\pi}{6}i}\) be a complex number written in polar form. Convert \(z\) to standard form, and write it in the form \(z = a+bi\).
If \(z\) and \(w\) are two complex numbers and the polar form of \(z\) involves the angle \(θ\) while the polar form of \(w\) involves the angle \(φ\), show that in the polar form for \(zw\) the angle involved is \(θ +φ\).
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You have \(z = |z|(\cos θ +i\sin θ)\) and \(w = |w|(\cos φ +i\sin φ)\). Then when you multiply these, you get \[\begin{aligned} &|z|\:|w| (\cos\theta +i\sin\theta )(\cos φ+i\sin φ) \\ =&|z|\:|w| (\cos\theta\cos φ-\sin\theta\sin φ+i(\cos\theta\sin φ+\cos φ\sin\theta )) \\ =&|z|\:|w| (\cos (\theta +φ)+i\sin (\theta+φ))\end{aligned}\]
Give the complete solution to \(x^4+16=0\).
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Solution is: \[(1-i)\sqrt{2},\: -(1+i)\sqrt{2},\: -(1-i)\sqrt{2},\: (1+i)\sqrt{2}\nonumber\]
Find the complex cube roots of \(8\).
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The cube roots are the solutions to \(z^3 - 8 = 0\), Solution is: \(-1 + i\sqrt{3},\: -1 + i\sqrt{3},\:2\)
Find the four fourth roots of \(-16\).
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The fourth roots are the solutions to \(z^4 + 16 = 0\), Solution is: \[(1-i)\sqrt{2},\:-(1+i)\sqrt{2},\:-(1-i)\sqrt{2},\:(1+i)\sqrt{2}\nonumber\]
De Moivre’s theorem says \([r(\cos t +i\sin t)]^n = r^n (\cos nt +i\sin nt)\) for \(n\) a positive integer. Does this formula continue to hold for all integers n, even negative integers? Explain.
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Yes, it holds for all integers. First of all, it clearly holds if \(n = 0\). Suppose now that n is a negative integer. Then \(−n > 0\) and so \[[r(\cos t+i\sin t)]^n=\frac{1}{[r(\cos t+i\sin t)]^{-n}}=\frac{1}{r^{-n}(\cos (-nt)+i\sin (-nt))}\nonumber\] \[\begin{aligned}&=\frac{r^n}{(\cos (nt)-i\sin (nt))}=\frac{r^n(\cos (nt)+i\sin (nt))}{(\cos (nt)-i\sin (nt))(\cos (nt)+i\sin (nt))} \\ &=r^n(\cos (nt)+i\sin (nt))\end{aligned}\] because \((\cos (nt)-i\sin (nt))(\cos (nt)+i\sin (nt))=1\).
Factor \(x^3 +8\) as a product of linear factors. Hint: Use the result of \(\PageIndex{14}\).
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Solution is: \(i\sqrt{3}+1,\: 1-i\sqrt{3},\: -2\) and so this polynomial equals \[(x+2)\left(x-\left(i\sqrt{3}+1\right)\right)\left(x-\left(1-i\sqrt{3}\right)\right)\nonumber\]
Write \(x^3 +27\) in the form \((x+3)(x^2 +ax+b)\) where \(x^2 +ax +b\) cannot be factored any more using only real numbers.
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\(x^3+27=(x+3)(x^2-3x+9)\)
Completely factor \(x^4 +16\) as a product of linear factors. Hint: Use the result of \(\PageIndex{15}\).
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Solution is: \[(1-i)\sqrt{2},\:-(1+i)\sqrt{2},\:-(1-i)\sqrt{2},\:(1+i)\sqrt{2}.\nonumber\] These are just the fourth roots of \(−16\). Then to factor, you get \[\left(x-\left((1-i)\sqrt{2}\right)\right)\left(x-\left(-(1+i)\sqrt{2}\right)\right).\nonumber\] \[\left(x-\left(-(1-i)\sqrt{2}\right)\right)\left(x-\left((1+i)\sqrt{2}\right)\right)\nonumber\]
Factor \(x^4 + 16\) as the product of two quadratic polynomials each of which cannot be factored further without using complex numbers.
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\(x^4+16=\left(x^2-2\sqrt{2}x+4\right)\left(x^2+2\sqrt{2}x+4\right)\). You can use the information in the preceding problem. Note that \((x−z) (x−\overline{z})\) has real coefficients.
If \(n\) is an integer, is it always true that \((\cos θ −i\sin θ)^n = \cos(nθ)−i\sin(nθ)\)? Explain.
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Yes, this is true. \[\begin{aligned}(\cos\theta -i\sin\theta)^n&=(\cos(-\theta)+i\sin(-\theta ))^n \\ &=\cos (-n\theta )+i\sin(-n\theta ) \\ &=\cos (n\theta )-i\sin (n\theta )\end{aligned}\]
Suppose \(p(x) = a_nx^n +a_{n−1}x^{n−1} +\cdots +a_1x+a_0\) is a polynomial and it has \(n\) zeros, \[z_1,\: z_2,\cdots ,z_n\nonumber\] listed according to multiplicity. (\(z\) is a root of multiplicity \(m\) if the polynomial \(f (x) = (x−z)^m\) divides \(p(x)\) but \((x−z) f (x)\) does not.) Show that \[p(x)=a_n(x-z_1)(x-z_2)\cdots (x-z_n)\nonumber\]
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\(p(x) = (x−z_1)q(x)+r(x)\) where \(r(x)\) is a nonzero constant or equal to \(0\). However, \(r(z_1) = 0\) and so \(r(x) = 0\). Now do to \(q(x)\) what was done to \(p(x)\) and continue until the degree of the resulting \(q(x)\) equals \(0\). Then you have the above factorization.
Show that \(1+i,\: 2+i\) are the only two roots to \[p(x) = x^2 −(3+2i)x+ (1+3i)\nonumber\] Hence complex zeros do not necessarily come in conjugate pairs if the coefficients of the equation are not real.
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\[(x−(1+i)) (x−(2+i)) = x^2 −(3+2i)x+1+3i\nonumber\]
Give the solutions to the following quadratic equations having real coefficients.
- \(x^2-2x+2=0\)
- \(3x^2+x+3=0\)
- \(x^2-6x+13=0\)
- \(x^2+4x+9=0\)
- \(4x^2+4x+5=0\)
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- Solution is: \(1+i,\: 1-i\)
- Solution is: \(\frac{1}{6}i\sqrt{35}-\frac{1}{6},\:-\frac{1}{6}i\sqrt{35}-\frac{1}{6}\)
- Solution is: \(3+2i,\: 3-2i\)
- Solution is: \(i\sqrt{5}-2,\:-i\sqrt{5}-2\)
- Solution is: \(-\frac{1}{2}+i,\:-\frac{1}{2}-i\)
Give the solutions to the following quadratic equations having complex coefficients.
- \(x^2+2x+1+i=0\)
- \(4x^2+4ix-5=0\)
- \(4x^2+(4+4i)x+1+2i=0\)
- \(x^2-4ix-5=0\)
- \(3x^2+(1-i)x+3i=0\)
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- Solution is: \(x=-1+\frac{1}{2}\sqrt{2}-\frac{1}{2}i\sqrt{2},\:x=-1-\frac{1}{2}\sqrt{2}+\frac{1}{2}i\sqrt{2}\)
- Solution is: \(x=1-\frac{1}{2}i,\:x=-1-\frac{1}{2}i\)
- Solution is: \(x=-\frac{1}{2},\:x=-\frac{1}{2}-i\)
- Solution is: \(x=-1+2i,\:x=1+2i\)
- Solution is: \(x=-\frac{1}{6}+\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}-\frac{1}{6}\sqrt{19}\right)i,\:x=-\frac{1}{6}-\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}+\frac{1}{6}\sqrt{19}\right)i\)
Prove the fundamental theorem of algebra for quadratic polynomials having coefficients in \(\mathbb{C}\). That is, show that an equation of the form \(ax^2 + bx + c = 0\) where \(a,\: b,\: c\) are complex numbers, \(a\neq 0\) has a complex solution. Hint: Consider the fact, noted earlier that the expressions given from the quadratic formula do in fact serve as solutions.