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Chapter 1 Exercise A


1.Solution: Because $(a+bi)(a-bi)=a^2+b^2$, one has\[\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}.\]Hence\[c=\frac{a}{a^2+b^2},d=-\frac{b}{a^2+b^2}.\]


2. Solution1:From direct computation, we have\[\left(\frac{-1+\sqrt{3}i}{2}\right)^2=\frac{-1-\sqrt{3}i}{2},\]hence \[\left(\frac{-1+\sqrt{3}i}{2}\right)^3=\frac{-1-\sqrt{3}i}{2}\cdot\frac{-1+\sqrt{3}i}{2}=1.\]This means $\dfrac{-1+\sqrt{3}i}{2}$ is a cube root of 1.


Solution2: Note that \[(a+bi)+(a-bi)=2a\] and \[(a+bi)(a-bi)=a^2+b^2,\] it follows that $\dfrac{-1+\sqrt{3}i}{2}$ is a root of the quadratic equation $x^2+x+1=0$.
For \[\frac{-1+\sqrt{3}i}{2}+\frac{-1-\sqrt{3}i}{2}=-1\] and \[\frac{-1+\sqrt{3}i}{2}\frac{-1-\sqrt{3}i}{2}=1.\] Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the conclusion.


3. Solution: If we know that $i=e^{\pi i/2}$, then the square roots are \[e^{\pi i/4}\quad\text{and}\quad e^{(\pi i/2+2\pi i)/2}=e^{5\pi i/4}.\] Note that for any $x\in\mathbb R$, one has $e^{xi}=\cos x+i\sin x$. Then \[e^{\pi i/4}=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{\sqrt{2}(1+i)}{2}\] and \[e^{5\pi i/4}=\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}=\frac{-\sqrt{2}(1+i)}{2}.\] Hence the roots are $\dfrac{\sqrt{2}(1+i)}{2}$ and $-\dfrac{\sqrt{2}(1+i)}{2}$.

Remark: If we don’t know this fact, then we should recall how to solve the roots of $x^8-1=0$ or $x^4+1=0$ since $x^2+i=0$ means $x^4+1=0$.


4. Solution: Let $\alpha=x+yi$ and $\beta=z+wi$, where $x,y,z,w\in\mathbb R$, then \[\alpha+\beta=(x+yi)+(z+wi)=(x+z)+(y+w)i.\] Similarly, \[\beta+\alpha=(z+wi)+(x+yi)=(z+x)+(w+y)i.\] Because $x+z=z+x$ and $y+w=w+y$, we obtain that $\alpha+\beta=\beta+\alpha$.


5. Solution: Let $\alpha=x_1+y_1i$, $\beta=x_2+y_2i$, $\lambda=x_3+y_3i$, where $x_1,x_2,x_3$ and $y_1,y_2,y_3$ are real numbers. Then \begin{aligned} (\alpha+\beta)+\lambda=&((x_1+x_2)+(y_1+y_2)i)+(x_3+y_3i)\\ =&((x_1+x_2)+x_3)+((y_1+y_2)+y_3)i. \end{aligned} Similarly, $\alpha+(\beta+\lambda)=(x_1+(x_2+x_3))+(y_1+(y_2+y_3))i$. Note that\[(x_1+x_2)+x_3=x_1+(x_2+x_3)\quad\text{and}\quad (y_1+y_2)+y_3=y_1+(y_2+y_3),\]it follows that $(\alpha+\beta)+\lambda=\alpha+(\beta+\lambda)$.


6. Solution: Let $\alpha=x_1+y_1i$, $\beta=x_2+y_2i$, $\lambda=x_3+y_3i$, where $x_1,x_2,x_3$ and $y_1,y_2,y_3$ are real numbers. Then \begin{align*} (\alpha\beta)\lambda=&((x_1x_2-y_1y_2)+(x_1y_2+y_1x_2)i)(x_3+y_3i)\\ =&((x_1x_2-y_1y_2)x_3-(x_1y_2+y_1x_2)y_3)\\&+((x_1x_2-y_1y_2)y_3+(x_1y_2+y_1x_2)x_3)i. \end{align*} Similarly, one has \begin{align*} \alpha(\beta\lambda)=&(x_1+y_1i)((x_2x_3-y_2y_3)+(x_2y_3+y_2x_3)i)\\ =&(x_1(x_2x_3-y_2y_3)-y_1(x_2y_3+y_2x_3))\\&+(x_1(x_2y_3+y_2x_3)+y_1(x_2x_3-y_2y_3))i. \end{align*} It is easy to see\[ (x_1x_2-y_1y_2)x_3-(x_1y_2+y_1x_2)y_3=x_1(x_2x_3-y_2y_3)-y_1(x_2y_3+y_2x_3) \]and\[ (x_1x_2-y_1y_2)y_3+(x_1y_2+y_1x_2)x_3=x_1(x_2y_3+y_2x_3)+y_1(x_2x_3-y_2y_3), \] hence we deduce that $(\alpha\beta)\lambda=\alpha(\beta\lambda)$.


7. Solution: Let $\alpha=x_1+y_1i$, $\beta=x_2+y_2i$, where $x_1,x_2$ and $y_1,y_2$ are real numbers. If $\alpha+\beta=0$, then \[ 0=\alpha+\beta=(x_1+x_2)+(y_1+y_2)i. \]This means $x_2=-x_1$ and $y_2=-y_1$, which implied uniqueness. If $\beta=-x_1-y_1i$, we also have $\alpha+\beta=0$, which implies existence.


8. Solution: We already know the existence in Problem 1. Now let us show the uniqueness, if $\alpha\beta=1$, then\[\beta=1\cdot\beta=\left(\frac{1}{\alpha}\cdot\alpha\right)\cdot\beta=\frac{1}{\alpha}\cdot\left(\alpha\cdot\beta\right)=\frac{1}{\alpha}\cdot1=\frac{1}{\alpha}.\]Here the third equality follows from Problem 6.


9. Solution: Suppose $\alpha=x_1+y_1i$, $\beta=x_2+y_2i$, $\lambda=a+bi$, where $x_1,x_2,a$ and $y_1,y_2,b$ are real numbers. Then \begin{align*} \lambda(\alpha+\beta)=&(a+bi)((x_1+x_2)+(y_1+y_2)i)\\ =&(a(x_1+x_2)-b(y_1+y_2))+(a(y_1+y_2)+b(x_1+x_2))i\\ =&[(ax_1-by_1)+(ay_1+bx_1)i]+[(ax_2-by_2)+(ay_2+bx_2)i]\\ =&\lambda\alpha+\lambda\beta. \end{align*}


10. Solution: Because $(4,-3,1,7)+2x=(5,9,-6,8)$, one has \[2x=(5,9,-6,8)-(4,-3,1,7)=(1,12,-7,1),\]hence\[x=\frac{1}{2}(1,12,-7,1)=\left(\frac{1}{2},6,\frac{-7}{2},\frac{1}{2}\right).\]


11. Solution: If such $\lambda\in\mathbb C$ exists, then we have \[\lambda(2-3i)=12-5i\quad\text{and}\quad\lambda(-6+7i)=-32-9i.\]It follows that \[(2-3i)(-32-9i)=(-6+7i)(12-5i),\]this means \[-91+78i=-37+114i, \]which is impossible. Hence such $\lambda\in\mathbb C$ does not exist.


12. Solution: Suppose $x=(x_1,\cdots,x_n)$, $y=(y_1,\cdots,y_n)$ and $z=(z_1,\cdots,z_n)$. Then \begin{align*} (x+y)+z=&((x_1,\cdots,x_n)+(y_1,\cdots,y_n))+(z_1,\cdots,z_n)\\ =&(x_1+y_1,\cdots,x_n+y_n)+(z_1,\cdots,z_n)\\ =&((x_1+y_1)+z_1,\cdots,(x_n+y_n)+z_n)\\ =&(x_1+(y_1+z_1),\cdots,x_n+(y_n+z_n))\\ =&(x_1,\cdots,x_n)+(y_1+z_1,\cdots,y_n+z_n)\\ =&(x_1,\cdots,x_n)+((y_1,\cdots,y_n)+(z_1,\cdots,z_n))\\ =&x+(y+z). \end{align*}


13. Solution: Suppose $x=(x_1,\cdots,x_n)$. Then \begin{align*} (ab)x=&ab(x_1,\cdots,x_n)=((ab)x_1,\cdots,(ab)x_n)\\ =&(a(bx_1),\cdots,a(bx_n))=a(bx_1,\cdots,bx_n)\\ =&a(bx). \end{align*}


14. Solution: Suppose $x=(x_1,\cdots,x_n)$. Then \[ 1x=1(x_1,\cdots,x_n)=(1\cdot x_1,\cdots,1\cdot x_n)=(x_1,\cdots,x_n)=x. \]


15. Solution: Suppose $x=(x_1,\cdots,x_n)$ and $y=(y_1,\cdots,y_n)$. Then \begin{align*} \lambda(x+y)=&\lambda((x_1,\cdots,x_n)+(y_1,\cdots,y_n))\\ =&\lambda(x_1+y_1,\cdots,x_n+y_n)=(\lambda(x_1+y_1),\cdots,\lambda(x_n+y_n))\\=&(\lambda x_1+\lambda y_1,\cdots,\lambda x_n+\lambda y_n)\\=&(\lambda x_1,\cdots,\lambda x_n)+(\lambda y_1,\cdots,\lambda y_n)\\ =&\lambda x+\lambda y. \end{align*}


16. Solution: Suppose $x=(x_1,\cdots,x_n)$. Then \begin{align*} (a+b)x=&(a+b)(x_1,\cdots,x_n)=((a+b)x_1,\cdots,(a+b)x_n)\\ =&(ax_1+bx_1,\cdots,ax_n+bx_n)\\ =&(ax_1,\cdots,ax_n)+(bx_1,\cdots,bx_n)\\ =&a(x_1,\cdots,x_n)+b(x_1,\cdots,x_n)\\ =&ax+bx. \end{align*}


Linearity

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This Post Has 12 Comments

  1. In the second solution of question 2, do the first 2 identities imply the solution of the quadratic equation?
    Thanks

  2. where can I download solution pdf file?

  3. Exercise 3: it must be (a+ib)^2=i
    a^2-b^2+2abi=i -> 1)a^2-b^2=0 and 2) 2ab=1
    From 1) we have a=b and a=-b. Substituting these equalities in 2) we get b=i/sqrt(2) and b=-i/sqrt(2)
    So the possible solution are (-1/sqrt(2)+i/sqrt(2)),
    (1/sqrt(2)-i/sqrt(2)), (1/sqrt(2)+i/sqrt(2)), (-1/sqrt(2)-i/sqrt(2)). But only the last two are those request.
    I think this proof is better than others because it uses only tha definition of complex number used in the theoretical part preceding this exercise.

    1. Sorry I did not see that someone else posted the same solution.

  4. Solution to exercise 2 can be improved by instead of solving the cube of (-1+i.3^0.5)/2, try finding a generic solution to (a+bi)^3 = 1. This will give 3 roots including 1 and (-1 +/- i.3^0.5)/2

  5. a typo in 15.
    for all lambdainmathbb F and all lambdainmathbb F^n
    should be
    for all lambdainmathbb F and all x,y inmathbb F^n

  6. N.11 isn't suppose to be -91+78i=-37+114i? On the LH: 2 * -32 = -64 ???

    1. Fixed, thanks!

  7. what a mistake?No.7,x2=-x1 is correct,but,y2=-x1 is wrong...it should be y2=-y1

    1. Fixed, thank you!

  8. For 3, let a and b be real numbers and (a+bi)^2 = i, then we have a^2-b^2=0 and 2ab=1, b=1/2a, a^2=b^2=1/4a^2, a^4=1/4, a^2=1/2, a=+-1/sqrt(2), b=+-1/sqrt(2), thus the solutions are +-1/sqrt(2)*(1+i).

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