Practice: Complex Numbers

(a) Product and quotient.

So $\mathrm{Re}(zw)=11$ and $\mathrm{Im}(zw)=10$, both real numbers. Modulus check. $|z|=\sqrt{4+9}=\sqrt{13}$, $|w|=\sqrt{16+1}=\sqrt{17}$, so $|z||w|=\sqrt{221}\approx 14.866$. And $|zw|=\sqrt{121+100}=\sqrt{221}$ ✓. For the quotient, $\overline{w}=4+i$ and $|w{|}^2=17$:

So $\mathrm{Re}(z/w)=5/17$ and $\mathrm{Im}(z/w)=14/17$. Check. $(\frac{5}{17}+\frac{14}{17}i)(4-i)=\frac{20-5i+56i-14i^2}{17}=\frac{20+51i+14}{17}=\frac{34+51i}{17}=2+3i=z$ ✓. (b) Polar form and a sixth power. $v=-1+i$ has $r=\sqrt{1+1}=\sqrt{2}$. The point lies in the second quadrant, negative real part, positive imaginary part, so the argument is $3\pi /4$, not the $-\pi /4$ a calculator's $\arctan (1/-1)=\arctan (-1)$ would return. This is the quadrant adjustment.

By De Moivre,

Reducing the argument modulo $2\pi$: $9\pi /2-4\pi =\pi /2$, so $v^6=8e^{i\pi /2}=8i$. Check. $v^2=(-1+i{)}^2=1-2i+i^2=-2i$, so $v^6=(v^2{)}^3=(-2i{)}^3=-8i^3=-8(-i)=8i$ ✓. (c) Cube roots of 8. Write $8=8e^{i\cdot 0}$. The roots have modulus $8^{1/3}=2$ and arguments $0$, $2\pi /3$, $4\pi /3$:

Verification. $(-1+\sqrt{3}i{)}^2=1-2\sqrt{3}i+3i^2=-2-2\sqrt{3}i$, and multiplying again by $(-1+\sqrt{3}i)$ gives $2-2\sqrt{3}i+2\sqrt{3}i-2\cdot 3i^2=2+6=8$ ✓. Sum. $2+(-1+\sqrt{3}i)+(-1-\sqrt{3}i)=0$ ✓. The two non-real roots are conjugates, so their imaginary parts cancel and their real parts $-1$ each cancel half of the 2. (d) The spectrum of a real matrix. $1-i$ is non-real, and the characteristic polynomial of a real matrix has real coefficients, so its conjugate $1+i$ must also be an eigenvalue, with the same algebraic multiplicity. That accounts for two of the four; the remaining two are either both real, or another conjugate pair. For a $5\times 5$ real matrix: the non-real eigenvalues pair off, so they occupy an even number of the five slots. An odd number therefore remains, and at least one eigenvalue must be real. This is why every rotation of ${\mathbb{R}}^3$ has an axis. A real eigenvector it leaves fixed. (e) Real coefficients, complex roots. The roots of $x^2-2x+5$ are $1\pm 2i$, and the coefficients are recovered from them:

Both are real, because a conjugate pair always sums to $2\mathrm{Re}$ and multiplies to $|\cdot {|}^2$. The imaginary parts cancel in the sum and the cross terms cancel in the product. So the polynomial $(x-\lambda )(x-\overline{\lambda })=x^2-2\mathrm{Re}(\lambda )x+|\lambda {|}^2$ has real coefficients even though neither factor does. The general principle. Conjugation respects addition and multiplication and fixes exactly the reals, so for a real-coefficient polynomial $\overline{p(\lambda )}=p(\overline{\lambda })$. A root is therefore carried to a root, and non-real roots come in pairs. The complex numbers appear in the working and cancel from anything the real world was asked about, which is the same reason a real differential equation with oscillating solutions is solved through $e^{i\omega t}$ and returns a real answer.