Exercises — Argand plane — geometric representation

WHAT: split $z=x+iy$ into $x=-2$, $y=3$. WHY: the real part is the horizontal coordinate, the imaginary part the vertical one. LOOKS LIKE: left of the origin (negative $x$) and above it (positive $y$) → Quadrant II. Point $(-2,3)$.

• $z_1=5+0i$: imaginary part is $0$, so it lies on the real axis at $(5,0)$.
• $z_2=0-4i$: real part is $0$, so it lies on the imaginary axis at $(0,-4)$.
• $z_3=0+0i$: the origin itself. Why: a zero part kills all motion along that axis, pinning the point to the other axis.

Legs of the right triangle are $6$ and $8$. $|z|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.$ Why Pythagoras and not something else? The modulus is a straight-line distance, and Pythagoras is exactly the tool that converts two perpendicular legs into a hypotenuse.

Step 1 (modulus): $r=\sqrt{1^2+(\sqrt{3}{)}^2}=\sqrt{1+3}=2$. Step 2 (reference angle): $\alpha ={\tan }^{-1}\left|\frac{\sqrt{3}}{1}\right|={\tan }^{-1}(\sqrt{3})=\frac{\pi }{3}$. Step 3 (quadrant fix): $x>0,y>0$ → Quadrant I → $\arg z=\alpha =\frac{\pi }{3}$. Step 4: $z=2\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$.

WHAT: $x=-\sqrt{3}<0$, $y=1>0$ → Quadrant II. Reference angle: $\alpha ={\tan }^{-1}\left|\frac{1}{\sqrt{3}}\right|=\frac{\pi }{6}$. Quadrant fix: in Q II, $\arg z=\pi -\alpha =\pi -\frac{\pi }{6}=\frac{5\pi }{6}$. Why not just ${\tan }^{-1}(1/(-\sqrt{3}))=-\pi /6$? Because $\tan$ repeats every $\pi$; the calculator lands in Q IV, but our arrow genuinely points up-and-left in Q II.

WHAT: $z_1-z_2=(4-1)+(2-(-2))i=3+4i$. WHY subtract first: $|z_1-z_2|$ is the straight-line distance between the two points, and subtraction gives the arrow from $P_2$ to $P_1$. $|z_1-z_2|=\sqrt{3^2+4^2}=5.$

WHAT: $x=-5$, $y=0$. The point sits on the negative real axis. WHY ${\tan }^{-1}$ dies here: $\tan \theta =y/x=0/(-5)=0$, which also holds for $\theta =0$ (positive real axis). The ratio can't tell the two apart. Read the picture: the arrow points straight left → angle $\pi$ from the positive real axis. $\arg (-5)=\pi .$ Note: $-\pi$ would be the same direction, but the principal range is $(-\pi ,\pi ]$, which includes $+\pi$ and excludes $-\pi$. So the answer is $\pi$.

For $z=3i$: $x=0$, so $\tan \theta =y/x$ is undefined (division by zero) — another warning that the formula alone can't cope. Look at the arrow: $3i$ points straight up → $\arg (3i)=\frac{\pi }{2}$. $-3i$ points straight down → $\arg (-3i)=-\frac{\pi }{2}$. Why not $+3\pi /2$ for the second? Anything past $\pi$ folds into the negative side to stay inside $(-\pi ,\pi ]$.

WHAT each side means: $|z-3|$ is the distance from $z$ to $(3,0)$; $|z+1|=|z-(-1)|$ is the distance from $z$ to $(-1,0)$. WHY: the equation says "equally far from $(3,0)$ and $(-1,0)$" — that is the definition of the perpendicular bisector of the segment joining them. Find it algebraically with $z=x+iy$: $(x-3{)}^2+y^2=(x+1{)}^2+y^2.$ Expand: $x^2-6x+9=x^2+2x+1\Rightarrow -6x+9=2x+1\Rightarrow 8=8x\Rightarrow x=1.$ Answer: the vertical line $x=1$ (midway between $x=3$ and $x=-1$, exactly as the geometry predicted).

Modulus: $r=\sqrt{(-2{)}^2+(-2{)}^2}=\sqrt{8}=2\sqrt{2}$. Reference angle: $\alpha ={\tan }^{-1}\left|\frac{-2}{-2}\right|={\tan }^{-1}(1)=\frac{\pi }{4}$. Quadrant fix: $x<0,y<0$ → Quadrant III → $\arg z=-(\pi -\alpha )=-\left(\pi -\frac{\pi }{4}\right)=-\frac{3\pi }{4}$. Polar form: $z=2\sqrt{2}\left(\cos \left(-\frac{3\pi }{4}\right)+i\sin \left(-\frac{3\pi }{4}\right)\right)$. Check by expanding: $\cos (-\frac{3\pi }{4})=-\frac{1}{\sqrt{2}}$, $\sin (-\frac{3\pi }{4})=-\frac{1}{\sqrt{2}}$, so $2\sqrt{2}\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\right)=-2-2i.✓$

Add tip-to-tail: $z_1+z_2=(2-1)+(3+1)i=1+4i$. Modulus of the sum: $|z_1+z_2|=\sqrt{1^2+4^2}=\sqrt{17}$. LOOKS LIKE: the diagonal of the parallelogram built on $\overrightarrow{O{P}_1}$ and $\overrightarrow{O{P}_2}$; its length is $\sqrt{17}$. Sanity check (triangle inequality): $|z_1|=\sqrt{13}\approx 3.61$, $|z_2|=\sqrt{2}\approx 1.41$, sum $\approx 5.02$, and $\sqrt{17}\approx 4.12\le 5.02$. ✓

WHAT: "distance from origin" $=|z|$; "distance from $(3,0)$" $=|z-3|$. "Twice as far from origin" means $|z|=2|z-3|$. WHY square: to remove the square roots cleanly, square both sides: $x^2+y^2=4((x-3{)}^2+y^2).$ Expand: $x^2+y^2=4x^2-24x+36+4y^2\Rightarrow 0=3x^2+3y^2-24x+36.$ Divide by 3: $x^2+y^2-8x+12=0\Rightarrow (x-4{)}^2+y^2=4.$ Answer: a circle, centre $(4,0)$, radius $2$ (an Apollonius circle).

WHAT is $i$ in polar form? $i=0+1i$ has $r=1$ and points straight up, so $i=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}$. Multiply: using $\cos A\cos B-\sin A\sin B=\cos (A+B)$ and $\sin A\cos B+\cos A\sin B=\sin (A+B)$, $i\cdot z=1\cdot r(\cos (\theta +\frac{\pi }{2})+i\sin (\theta +\frac{\pi }{2})).$ Read it: the length $r$ is unchanged (factor of $1$), while the angle grew by $\frac{\pi }{2}$. That is exactly a $90^{\circ }$ anticlockwise turn. See Multiplication as Rotation and Scaling. Concrete check: $i(2+3i)=2i+3i^2=-3+2i$; indeed $(2,3)$ rotated $90^{\circ }$ anticlockwise lands at $(-3,2)$. ✓

Polar form of $1+i$: $r=\sqrt{2}$, Quadrant I, $\alpha ={\tan }^{-1}(1)=\frac{\pi }{4}$, so $1+i=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$. Why polar? Powers scramble in rectangular form but obey a clean rule in polar: raise the length, multiply the angle — that is De Moivre's Theorem. $(1+i{)}^8=(\sqrt{2}{)}^8(\cos (8\cdot \frac{\pi }{4})+i\sin (8\cdot \frac{\pi }{4}))=16(\cos 2\pi +i\sin 2\pi ).$ Since $\cos 2\pi =1$, $\sin 2\pi =0$: $(1+i{)}^8=16$.

Rewrite: $|z-1|=|z+1|$ (multiply both sides by $|z+1|$, valid whenever $z\ne -1$). Geometry: distance from $(1,0)$ equals distance from $(-1,0)$ → perpendicular bisector of that segment → the imaginary axis $x=0$. Algebra: $(x-1{)}^2+y^2=(x+1{)}^2+y^2\Rightarrow -2x=2x\Rightarrow x=0.$ ✓ Caveat: $z=-1$ is excluded (it would divide by zero), but $z=-1$ has $x=-1\ne 0$ anyway, so it was never on the line — no points lost. Answer: the imaginary axis. See Loci in the Complex Plane.

Recognise the point and its axes (L1) → compute modulus/argument/distance (L2) → handle every degenerate axis case and the argument's half-open range (L3) → synthesise polar form, addition, and word-problems into equations (L4) → prove rotation, wield De Moivre, and decode ratio-loci (L5).