Exercises — Algebraic operations — add, subtract, multiply, divide (rectangular and polar)

WHAT: add two arrows tip-to-tail. WHY rectangular: addition just piles up horizontal reach and vertical reach separately (look at s01, the two arrows stack into the dashed resultant). $(4+3i)+(2-7i)=(4+2)+(3-7)i=6-4i.$ Real parts $4,2$ add; imaginary parts $3,-7$ add. Never mix the axes.

Subtraction = adding the reversed arrow. $(4+3i)-(2-7i)=(4-2)+(3-(-7))i=2+10i.$ Watch the double sign: $3-(-7)=10$.

WHY these tools: $r=\sqrt{a^2+b^2}$ is the arrow length (Pythagoras on the shadows $a=1,b=\sqrt{3}$); the argument answers "which angle points this way?" $r=\sqrt{1^2+(\sqrt{3}{)}^2}=\sqrt{1+3}=\sqrt{4}=2.$ Both $a>0$ and $b>0$ → quadrant I, so plain arctan is safe here: $\theta =\arctan \left(\frac{\sqrt{3}}{1}\right)=60^{\circ }.$ $z=2(\cos 60^{\circ }+i\sin 60^{\circ }).$

WHAT: FOIL, then use $i^2=-1$. WHY the sign flip: the $bd{i}^2$ term becomes $-bd$ and drops into the real part. $(3+2i)(1-4i)=3\cdot 1+3(-4i)+2i\cdot 1+2i(-4i)$ $=3-12i+2i-8i^2=3-10i+8=11-10i.$ Here $-8i^2=-8(-1)=+8$, added to the real $3$.

WHY the conjugate: we cannot divide by an arrow. Multiplying top and bottom by $\bar{z}=3+2i$ makes the denominator real, because $(3-2i)(3+2i)=3^2+2^2=13$. $\frac{2+i}{3-2i}\cdot \frac{3+2i}{3+2i}=\frac{(2+i)(3+2i)}{13}.$ Numerator: $6+4i+3i+2i^2=6+7i-2=4+7i$. $=\frac{4+7i}{13}=\frac{4}{13}+\frac{7}{13}i.$

WHY polar: multiplication scales length and rotates angle (look at s02: the product arrow is longer and swung further round). $z_1{z}_2=(2\cdot 3)[\cos (40^{\circ }+25^{\circ })+i\sin (40^{\circ }+25^{\circ })]=6(\cos 65^{\circ }+i\sin 65^{\circ }).$ Moduli multiply ($2\cdot 3=6$), arguments add ($40^{\circ }+25^{\circ }=65^{\circ }$).

Rectangular: $(-1-i{)}^2=(-1{)}^2+2(-1)(-i)+(-i{)}^2=1+2i+i^2=1+2i-1=2i.$ Polar: first place $-1-i$. Both shadows negative → quadrant III (look at s01, lower-left). Length $r=\sqrt{(-1{)}^2+(-1{)}^2}=\sqrt{2}$. Naive $\arctan \frac{-1}{-1}=\arctan (1)=45^{\circ }$ is wrong — that points into quadrant I. Since $a<0$ we add $180^{\circ }$: $\theta =45^{\circ }+180^{\circ }=225^{\circ }$. Square it: modulus $(\sqrt{2}{)}^2=2$, argument $2\times 225^{\circ }=450^{\circ }\equiv 90^{\circ }$. $=2(\cos 90^{\circ }+i\sin 90^{\circ })=2(0+i\cdot 1)=2i.✓$ Both routes give $2i$.

WHAT: solve for $z$, i.e. $z=\frac{5}{2-i}$. WHY: dividing undoes the multiplication. $z=\frac{5}{2-i}\cdot \frac{2+i}{2+i}=\frac{5(2+i)}{4+1}=\frac{10+5i}{5}=2+i.$ Check: $(2-i)(2+i)=4+1=5.✓$

WHY polar: reciprocal = divide $1$ (modulus $1$, argument $0^{\circ }$) by $z$, so divide moduli and subtract arguments. $\frac{1}{z}=\frac{1}{4}[\cos (0^{\circ }-150^{\circ })+i\sin (0^{\circ }-150^{\circ })]=\frac{1}{4}[\cos (-150^{\circ })+i\sin (-150^{\circ })].$ $-150^{\circ }$ is already in the required range.

Step A (polar, WHY): raising to the 4th power means 4 rotations + 4 stretches — trivial in polar. $1+i$ has $r=\sqrt{2}$, $\theta =45^{\circ }$ (quadrant I). Then $(1+i{)}^4=(\sqrt{2}{)}^4[\cos (4\cdot 45^{\circ })+i\sin (4\cdot 45^{\circ })]=4(\cos 180^{\circ }+i\sin 180^{\circ })=4(-1)=-4.$ Step B (rectangular division): $\frac{-4}{2i}=\frac{-4}{2i}\cdot \frac{-i}{-i}=\frac{4i}{-2i^2}=\frac{4i}{2}=2i.$ Here multiplying by $\frac{-i}{-i}=1$ clears the imaginary denominator (since $2i\cdot (-i)=-2i^2=2$). $\boxed{\frac{(1+i{)}^4}{2i}=2i.}$

Here $\bar{z}=1-i$ (flip the vertical shadow — mirror across the real axis).

• $z+\bar{z}=(1+i)+(1-i)=2=2\mathrm{Re}(z)$. Sum kills the imaginary part → twice the real part.
• $z-\bar{z}=(1+i)-(1-i)=2i=2i\mathrm{Im}(z)$. Difference kills the real part.
• $z\bar{z}=(1+i)(1-i)=1-i^2=2=|z{|}^2$. Product of a number with its conjugate is the squared length, always real and non-negative.

Proof (polar, WHY it is one line): write $z_1=r_1{e}^{i\alpha }$, $z_2=r_2{e}^{i\beta }$. Then $z_1{z}_2=r_1{r}_2{e}^{i(\alpha +\beta )}.$ The modulus is the coefficient in front of $e^{i(\cdot )}$ (since $|e^{i\varphi }|=1$ for every angle $\varphi$ — it is a unit-length arrow). Hence $|z_1{z}_2|=r_1{r}_2=|z_1||z_2|.■$ Numeric check: $|z_1|=\sqrt{3^2+4^2}=5$, $|z_2|=\sqrt{1^2+2^2}=\sqrt{5}$. Product: $(3+4i)(1+2i)=3+6i+4i+8i^2=3+10i-8=-5+10i$, with modulus $\sqrt{(-5{)}^2+10^2}=\sqrt{125}=5\sqrt{5}$. And $|z_1||z_2|=5\cdot \sqrt{5}=5\sqrt{5}.✓$

WHY: $i$ in polar form is $1\cdot (\cos 90^{\circ }+i\sin 90^{\circ })=e^{i90^{\circ }}$ — a unit arrow (modulus $1$) pointing straight up. Multiplying by it therefore scales length by $1$ (no stretch) and adds $90^{\circ }$ to the angle (pure rotation). Demonstration (rectangular): $i(3+2i)=3i+2i^2=-2+3i.$ Check the rotation on shadows: $(3,2)\to (-2,3)$ is exactly the $90^{\circ }$ counter-clockwise turn $(x,y)\to (-y,x)$. Length unchanged: $\sqrt{3^2+2^2}=\sqrt{13}=\sqrt{(-2{)}^2+3^2}.✓$ (Look at s02 — the coral arrow is the lavender one swung a quarter-turn.)

WHY polar: taking a square root = halving the angle and rooting the length — and a full circle offers two starting angles ($\theta$ and $\theta +360^{\circ }$) that both halve to valid roots. Write $2i$ in polar: modulus $2$, argument $90^{\circ }$. So $2i=2(\cos 90^{\circ }+i\sin 90^{\circ })$, and also $=2(\cos 450^{\circ }+i\sin 450^{\circ })$ (one extra turn). A root has modulus $\sqrt{2}$ and argument half of each: $z_1=\sqrt{2}(\cos 45^{\circ }+i\sin 45^{\circ })=\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)=1+i,$ $z_2=\sqrt{2}(\cos 225^{\circ }+i\sin 225^{\circ })=\sqrt{2}\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\right)=-1-i.$ Check: $(1+i{)}^2=2i$ and $(-1-i{)}^2=2i$ (from Problem 7). ✓ The two roots are $\pm (1+i)$ — opposite arrows, $180^{\circ }$ apart, as square roots always are.

You practised one habit: pick the form that matches the operation. Piling arrows up (add/subtract) → stay in the box (rectangular). Stretch-and-spin jobs (multiply, divide, power, root) → jump onto the clock (polar). Divisions in rectangular need the conjugate to make the bottom real; roots on the clock need you to remember there's more than one answer hiding a full turn away.