# 6.E: Exercises

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## Exercise $$\PageIndex{1}$$

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

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

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.

$\overline{(a+bi) (c+di)} = \overline{ac−bd + (ad +bc)i} = (ac−bd)−(ad +bc)i(a−bi) (c−di) = ac−bd −(ad +bc)i\nonumber$ 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.

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?

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 $$θ +φ$$.

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 $$x^4+16=0$$.

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 $$8$$.

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 $$-16$$.

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.

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 $$\PageIndex{14}$$.

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.

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

## Exercise $$\PageIndex{19}$$

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

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.

$$x^4+16=\left(x^2-2\sqrt{2}x+4\right)\left(x^2+2\sqrt{2}x+4\right)$$. 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.

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

$$p(x) = (x−z_1)q(x)+r(x)$$ 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.

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

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

This page titled 6.E: Exercises is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .