6.E: Exercises

Exercise $\PageIndex{1}$

Let $z = 2+7i$ and let $w = 3−8i$. Compute the following.

1. $z+w$
2. $z-2w$
3. $zw$
4. $\frac{w}{z}$

Answer

1. $z+w=5-i$
2. $z-2w=-4+23i$
3. $zw=62+5i$
4. $\frac{w}{z}=-\frac{50}{53}-\frac{37}{53}i$

Exercise $\PageIndex{2}$

Let $z = 1−4i$. Compute the following.

1. $\overline{z}$
2. $z^{-1}$
3. $|z|$

Exercise $\PageIndex{3}$

Let $z = 3+5i$ and $w = 2−i$. Compute the following.

1. $\overline{zw}$
2. $|zw|$
3. $z^{-1}w$

Exercise $\PageIndex{4}$

If $z$ is a complex number, show there exists a complex number $w$ with $|w| = 1$ and $wz = |z|$.

Answer

If $z=0$, let $w=1$. If $z\neq 0$, let $w=\frac{\overline{z}}{|z|}$

Exercise $\PageIndex{5}$

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.

Answer

$$\overline{(a+bi) (c+di)} = \overline{ac−bd + (ad +bc)i} = (ac−bd)−(ad +bc)i(a−bi) (c−di) = ac−bd −(ad +bc)i\nonumber$$ 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. $$\overline{\sum\limits_{j=1}^n(x_j+iy_j)}=\overline{\sum\limits_{j=1}^nx_j+i\sum\limits_{j=1}^ny_j}\nonumber$$ $$=\sum\limits_{j=1}^nx_j-i\sum\limits_{j=1}^ny_j=\sum\limits_{j=1}^nx_j-iy_j=\sum\limits_{j=1}^n\overline{(x_j+iy_j)}.\nonumber$$

Exercise $\PageIndex{6}$

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.

Answer

If $p(z)=0$, then you have $$\begin{aligned}\overline{p(z)}&=0=\overline{a_nz^n+a_{n-1}z^{n-1}+\cdots +a_1z+a_0} \\ &=\overline{a_nz^n}+\overline{a_{n-1}z^{n-1}}+\cdots +\overline{a_1z}+\overline{a_0} \\ &=\overline{a_n}\:\overline{z}^n+\overline{a_{n-1}}\:\overline{z}^{n-1}+\cdots +\overline{a_1}\:\overline{z}+\overline{a_0} \\ &=a_n\overline{z}^n+a_{n-1}\overline{z}^{n-1}+\cdots +a_1\overline{z}+a_0 \\ &=p(\overline{z})\end{aligned}$$

Exercise $\PageIndex{7}$

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?

Answer

The problem is that there is no single $\sqrt{-1}$.

Exercise $\PageIndex{8}$

Let $z = 3+3i$ be a complex number written in standard form. Convert $z$ to polar form, and write it in the form $z = re^{iθ}$.

Exercise $\PageIndex{9}$

Let $z = 2i$ be a complex number written in standard form. Convert $z$ to polar form, and write it in the form $z = re^{iθ}$.

Exercise $\PageIndex{10}$

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 $z = a+bi$.

Exercise $\PageIndex{11}$

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 $z = a+bi$.

Exercise $\PageIndex{12}$

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 $θ +φ$.

Answer

You have $z = |z|(\cos θ +i\sin θ)$ and $w = |w|(\cos φ +i\sin φ)$. Then when you multiply these, you get $$\begin{aligned} &|z|\:|w| (\cos\theta +i\sin\theta )(\cos φ+i\sin φ) \\ =&|z|\:|w| (\cos\theta\cos φ-\sin\theta\sin φ+i(\cos\theta\sin φ+\cos φ\sin\theta )) \\ =&|z|\:|w| (\cos (\theta +φ)+i\sin (\theta+φ))\end{aligned}$$

Exercise $\PageIndex{13}$

Give the complete solution to $x^4+16=0$.

Answer

Solution is: $$(1-i)\sqrt{2},\: -(1+i)\sqrt{2},\: -(1-i)\sqrt{2},\: (1+i)\sqrt{2}\nonumber$$

Exercise $\PageIndex{14}$

Find the complex cube roots of $8$.

Answer

The cube roots are the solutions to $z^3 - 8 = 0$, Solution is: $-1 + i\sqrt{3},\: -1 + i\sqrt{3},\:2$

Exercise $\PageIndex{15}$

Find the four fourth roots of $-16$.

Answer

The fourth roots are the solutions to $z^4 + 16 = 0$, Solution is: $$(1-i)\sqrt{2},\:-(1+i)\sqrt{2},\:-(1-i)\sqrt{2},\:(1+i)\sqrt{2}\nonumber$$

Exercise $\PageIndex{16}$

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.

Answer

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 $(\cos (nt)-i\sin (nt))(\cos (nt)+i\sin (nt))=1$.

Exercise $\PageIndex{17}$

Factor $x^3 +8$ as a product of linear factors. Hint: Use the result of $\PageIndex{14}$.

Answer

Solution is: $i\sqrt{3}+1,\: 1-i\sqrt{3},\: -2$ and so this polynomial equals $$(x+2)\left(x-\left(i\sqrt{3}+1\right)\right)\left(x-\left(1-i\sqrt{3}\right)\right)\nonumber$$

Exercise $\PageIndex{18}$

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.

Answer

$x^3+27=(x+3)(x^2-3x+9)$

Exercise $\PageIndex{19}$

Completely factor $x^4 +16$ as a product of linear factors. Hint: Use the result of $\PageIndex{15}$.

Answer

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 $$\left(x-\left((1-i)\sqrt{2}\right)\right)\left(x-\left(-(1+i)\sqrt{2}\right)\right).\nonumber$$ $$\left(x-\left(-(1-i)\sqrt{2}\right)\right)\left(x-\left((1+i)\sqrt{2}\right)\right)\nonumber$$

Exercise $\PageIndex{20}$

Factor $x^4 + 16$ as the product of two quadratic polynomials each of which cannot be factored further without using complex numbers.

Answer

$x^4+16=\left(x^2-2\sqrt{2}x+4\right)\left(x^2+2\sqrt{2}x+4\right)$. You can use the information in the preceding problem. Note that $(x−z) (x−\overline{z})$ has real coefficients.

Exercise $\PageIndex{21}$

If $n$ is an integer, is it always true that $(\cos θ −i\sin θ)^n = \cos(nθ)−i\sin(nθ)$? Explain.

Answer

Yes, this is true. $$\begin{aligned}(\cos\theta -i\sin\theta)^n&=(\cos(-\theta)+i\sin(-\theta ))^n \\ &=\cos (-n\theta )+i\sin(-n\theta ) \\ &=\cos (n\theta )-i\sin (n\theta )\end{aligned}$$

Exercise $\PageIndex{22}$

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 $$p(x)=a_n(x-z_1)(x-z_2)\cdots (x-z_n)\nonumber$$

Answer

$p(x) = (x−z_1)q(x)+r(x)$ 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.

Exercise $\PageIndex{23}$

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.

Answer

$$(x−(1+i)) (x−(2+i)) = x^2 −(3+2i)x+1+3i\nonumber$$

Exercise $\PageIndex{24}$

Give the solutions to the following quadratic equations having real coefficients.

1. $x^2-2x+2=0$
2. $3x^2+x+3=0$
3. $x^2-6x+13=0$
4. $x^2+4x+9=0$
5. $4x^2+4x+5=0$

Answer

1. Solution is: $1+i,\: 1-i$
2. Solution is: $\frac{1}{6}i\sqrt{35}-\frac{1}{6},\:-\frac{1}{6}i\sqrt{35}-\frac{1}{6}$
3. Solution is: $3+2i,\: 3-2i$
4. Solution is: $i\sqrt{5}-2,\:-i\sqrt{5}-2$
5. Solution is: $-\frac{1}{2}+i,\:-\frac{1}{2}-i$

Exercise $\PageIndex{25}$

Give the solutions to the following quadratic equations having complex coefficients.

1. $x^2+2x+1+i=0$
2. $4x^2+4ix-5=0$
3. $4x^2+(4+4i)x+1+2i=0$
4. $x^2-4ix-5=0$
5. $3x^2+(1-i)x+3i=0$

Answer

1. Solution is: $x=-1+\frac{1}{2}\sqrt{2}-\frac{1}{2}i\sqrt{2},\:x=-1-\frac{1}{2}\sqrt{2}+\frac{1}{2}i\sqrt{2}$
2. Solution is: $x=1-\frac{1}{2}i,\:x=-1-\frac{1}{2}i$
3. Solution is: $x=-\frac{1}{2},\:x=-\frac{1}{2}-i$
4. Solution is: $x=-1+2i,\:x=1+2i$
5. Solution is: $x=-\frac{1}{6}+\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}-\frac{1}{6}\sqrt{19}\right)i,\:x=-\frac{1}{6}-\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}+\frac{1}{6}\sqrt{19}\right)i$

Exercise $\PageIndex{26}$

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.