Modulus, Argument & Conjugate

$z=a+ib$, $a=\operatorname{Re}(z)$, $b=\operatorname{Im}(z)$, $i^2=-1$. Equality: $a+ib=c+id\iff a=c$ and $b=d$. Arithmetic: add/subtract componentwise; multiply as binomials ($(a+ib)(c+id)=(ac-bd)+i(ad+bc)$). Divide by multiplying top and bottom by the denominator's conjugate.

• The $1\pm i$ shortcuts — eight PYQs open with one of these: $(1+i)^2=2i$, $(1-i)^2=-2i$, $|1\pm i|=\sqrt2$, $\dfrac{1+i}{1-i}=i$, $\dfrac{1-i}{1+i}=-i$. So $\Big(\dfrac{1+i}{1-i}\Big)^n=i^n$ (smallest $n$ making it 1 is 4, making it $-1$ is 2), $(1+i)^4=(2i)^2=-4$, $(1+i)^{2n}=(2i)^n$.
• Complex numbers are NOT ordered. $<$ and $>$ are defined only for real numbers; a statement like $1+4i>3+2i$ is not false — it is meaningless, and every inequality between two non-real complex numbers is. Compare their MODULI if a question wants a size comparison.

$\overline{a+ib}=a-ib$. Key facts:

• $z\bar z=|z|^2=a^2+b^2$; $z+\bar z=2\operatorname{Re}(z)$; $z-\bar z=2i\operatorname{Im}(z)$.
• Purely real $\iff z=\bar z$ (imaginary part 0). Purely imaginary $\iff z=-\bar z$ (real part 0).
• $\overline{z_1z_2}=\bar z_1\bar z_2$, $\overline{z_1/z_2}=\bar z_1/\bar z_2$. A real-coefficient equation has complex roots in conjugate pairs.
• Reciprocal via the conjugate: $z^{-1}=\dfrac{\bar z}{|z|^2}$; in particular $|z|=1\Rightarrow z^{-1}=\bar z$.
• Argument of the conjugate: $\arg\bar z=-\arg z$ (reflection in the real axis), so $\operatorname{amp}(z)+\operatorname{amp}(\bar z)=0$. It is NOT $\pi-\arg z$ — that is the argument of $-\bar z$.

$|z|=\sqrt{a^2+b^2}$. Properties: $|z_1z_2|=|z_1||z_2|$, $\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}$, $|z^n|=|z|^n$, $|\bar z|=|z|$. Triangle inequality: $\big||z_1|-|z_2|\big|\le|z_1\pm z_2|\le|z_1|+|z_2|$ — gives the max/min of $|z\pm c|$ on a disc $|z-a|\le r$.

• Perpendicularity: $|z_1+z_2|=|z_1-z_2|\iff\dfrac{z_1}{z_2}$ is purely imaginary $\iff\operatorname{Re}(z_1\bar z_2)=0$ — the diagonals of the parallelogram on $z_1,z_2$ are equal, so it is a rectangle and the vectors are perpendicular ($\arg z_1-\arg z_2=\pm\tfrac\pi2$). Conversely $\dfrac{z_1}{z_2}$ purely imaginary gives $\Big|\dfrac{z_1+z_2}{z_1-z_2}\Big|=1$.
• Unit-modulus (Blaschke) identity: if $|\alpha|=1$ then $\left|\dfrac{\alpha-\beta}{1-\bar\alpha\beta}\right|=1$ for every $\beta\ne\alpha$ — because $1-\bar\alpha\beta=\bar\alpha(\alpha-\beta)$ when $\alpha\bar\alpha=1$, and $|\bar\alpha|=1$. The same holds with the roles of $\alpha,\beta$ swapped when $|\beta|=1$.

Polar form: $z=r(\cos\theta+i\sin\theta)=re^{i\theta}$, $r=|z|$, $\theta=\arg z$. The principal argument lies in $(-\pi,\pi]$; compute $\tan^{-1}\big|\tfrac{b}{a}\big|$ then adjust for the quadrant of $(a,b)$. Arguments add under multiplication: $\arg(z_1z_2)=\arg z_1+\arg z_2$, $\arg(z_1/z_2)=\arg z_1-\arg z_2$. Two conventions are in use. Textbooks fix the principal value in $(-\pi,\pi]$, but some NDA papers write it in $[0,2\pi)$ — the same angle read as $-\tfrac{2\pi}{3}$ or as $240^\circ$. The two differ by exactly $2\pi$ ($360^\circ$), so if your answer is missing from the options, add $2\pi$ before deciding you are wrong. The option list tells you which convention the setter used.

Recognise first, substitute second:

• $|z-a|=r$: a circle, centre $a$, radius $r$. $|z|=r$ is centred at the origin; $z\bar z=k$ is $x^2+y^2=k$.
• $|z-a|=|z-b|$: the perpendicular bisector of the segment $ab$ (a straight line). $|z-a|=k|z-b|$ with $k\ne1$ is a circle (Apollonius).
• $\operatorname{Re}(z^2)=c$, i.e. $x^2-y^2=c$: a rectangular hyperbola; $\operatorname{Im}(z^2)=c$, i.e. $2xy=c$, is the other rectangular hyperbola.
• $\arg(z-a)=\theta$: a ray from $a$; $\arg\dfrac{z-a}{z-b}=\pm\tfrac\pi2$: the circle on diameter $ab$ (angle in a semicircle).
• $\left|\dfrac{z-a}{z-b}\right|=1$: again the perpendicular bisector of $ab$ — so $\left|\dfrac{z+\lambda i}{z-\lambda i}\right|=1$ is the real axis $y=0$.

Substitute when nothing matches: $z\bar z=|z+\bar z|$ becomes $x^2+y^2=2|x|$, a pair of circles $(x\mp1)^2+y^2=1$; $z\bar z+|z|^2=4$ is $2(x^2+y^2)=4$, the circle $x^2+y^2=2$.