In the real numbers the equation $x^2=-1$ has no solution: the square of a real number is never negative. Complex numbers extend the real numbers so that square roots of negative numbers exist. On the YÖS exam (MAT-2) this subtopic gives fast, pattern-based questions: powers of $i$, real and imaginary parts, the conjugate $\bar z$, the modulus $|z|$, powers of binomials such as $(1+i)^n$, and the trigonometric form $r\,\text{cis}\,\theta$. Almost every problem reduces to a small set of rules and shortcuts collected below.
The imaginary unit is defined by$$i^2=-1,\qquad i=\sqrt{-1}.$$For a real number $a>0$: $\sqrt{-a}=i\sqrt{a}$. For example, $\sqrt{-36}+\sqrt{-25}=6i+5i=11i$.
A complex number is $z=a+bi$ with real $a,b$. Here $\text{Re}(z)=a$ is the real part and $\text{Im}(z)=b$ is the imaginary part (the coefficient $b$, not $bi$). Two complex numbers are equal iff their real parts and imaginary parts are equal: $a+bi=c+di \iff a=c$ and $b=d$.
Warning: the rule $\sqrt{a}\sqrt{b}=\sqrt{ab}$ fails for two negative radicands: $\sqrt{-4}\cdot\sqrt{-9}=(2i)(3i)=6i^2=-6$, not $+6$.
Negative radicand trick: if $x<y<0$, then $x-y<0$, so $\sqrt{x-y}=i\sqrt{y-x}$ is purely imaginary. For instance $\sqrt{x-y}=2i$ means $x-y=(2i)^2=-4$, i.e. $y-x=4$.
The powers of $i$ repeat with period 4:$$i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1,\quad i^5=i,\ \dots$$So $i^n$ depends only on the remainder of $n$ modulo 4:$$i^n=i^{\,n \bmod 4},\qquad i^0=1.$$Example: $143=4\cdot 35+3$, so $i^{143}=i^3=-i$; $172=4\cdot 43$, so $i^{172}=1$; hence $i^{143}+i^{172}=1-i$.
Key consequence: any four consecutive powers cancel,$$i^k+i^{k+1}+i^{k+2}+i^{k+3}=0.$$Therefore in a sum $\sum_{k=1}^{n} i^k$ only the leftover terms after full blocks of 4 survive. Example: $\sum_{k=1}^{73} i^k$ — since $73=4\cdot 18+1$, the first $72$ terms cancel and the sum equals $i^{73}=i$.
Weighted sums such as $i^2+2i^5+3i^8+\dots$: first reduce every exponent mod 4 ($i^2=-1$, $i^5=i$, $i^8=1$, $i^{11}=-i$, ...), then the values repeat with period 4 while the coefficients grow by 1; group the terms in blocks of four and count the number of full blocks and leftovers.
Operations are performed componentwise or by expanding brackets with $i^2=-1$:$$(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i,$$$$k(a+bi)=ka+kbi\ (k\in\mathbb{R}),$$$$(a+bi)(c+di)=(ac-bd)+(ad+bc)i.$$Typical YÖS pattern: given $z_1$ and $z_2$, compute a combination such as $2z_1-z_2$ and write it as $a+bi$, then find $a\cdot b$ or $a+b$. Example: $z_1=3+2i$, $z_2=1-i$ give $2z_1-z_2=(6+4i)-(1-i)=5+5i$, so $a\cdot b=25$.
After any computation always collect the answer into the standard form $a+bi$ before reading off $\text{Re}$, $\text{Im}$ or comparing coefficients.
The conjugate of $z=a+bi$ is $\bar z=a-bi$ (the sign of the imaginary part flips). Properties:$$\overline{z_1\pm z_2}=\bar z_1\pm\bar z_2,\qquad \overline{z_1 z_2}=\bar z_1\,\bar z_2,\qquad \overline{\left(\tfrac{z_1}{z_2}\right)}=\tfrac{\bar z_1}{\bar z_2},\qquad \bar{\bar z}=z.$$$$z+\bar z=2\,\text{Re}(z),\qquad z-\bar z=2i\,\text{Im}(z),\qquad z\bar z=a^2+b^2=|z|^2.$$Consequences: $z$ is real $\iff z=\bar z$; $z$ is purely imaginary $\iff z=-\bar z$ (and $z\ne 0$). The product $z\bar z$ is always a nonnegative real number — this is exactly what makes division possible.
To divide, multiply numerator and denominator by the conjugate of the denominator:$$\frac{z_1}{z_2}=\frac{z_1\,\bar z_2}{z_2\,\bar z_2}=\frac{z_1\,\bar z_2}{|z_2|^2}.$$The denominator becomes the real number $|z_2|^2$, and the result can be split into $\text{Re}$ and $\text{Im}$.
Useful shortcuts:$$\frac{1}{i}=-i,\qquad \frac{1+i}{1-i}=i,\qquad \frac{1-i}{1+i}=-i.$$Strategy for expressions like $z=\dfrac{(1+2i)(1-i)}{3-i}$ or $\dfrac{i}{(1-i)(2+i)}$: first multiply out the products in the numerator (or denominator) to get a single fraction $\frac{p+qi}{c+di}$, then multiply by $\frac{c-di}{c-di}$ and read off $\text{Re}(z)$ and $\text{Im}(z)$.
The modulus of $z=a+bi$ is the distance from the origin to the point $(a,b)$:$$|z|=\sqrt{a^2+b^2}.$$Properties (memorize — they save minutes on the exam):$$|z|=|\bar z|=|-z|,\qquad |z_1z_2|=|z_1|\,|z_2|,\qquad \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|},\qquad |z^n|=|z|^n,\qquad z\bar z=|z|^2.$$Classic YÖS trick: for real $x$,$$\left|\frac{1+ix}{1-ix}\right|=\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}=1$$— no need to simplify the fraction: numerator and denominator are conjugates, so they have equal moduli.
For expressions like $|z_1-2z_2|$ with given $z_1,z_2$: first compute the complex number $z_1-2z_2=a+bi$, then take $\sqrt{a^2+b^2}$. The modulus of a sum is NOT the sum of moduli; only the triangle inequality $|z_1+z_2|\le|z_1|+|z_2|$ holds.
Universal method: substitute $z=x+yi$, $\bar z=x-yi$, expand, and collect the equation into the form $A+Bi=C+Di$. Then equate parts: $A=C$ and $B=D$ — one complex equation gives a system of two real equations in $x,y$.
Example setup for $i\bar z+1=2(1-z)$: since $i\bar z=i(x-yi)=y+xi$, the left side is $(y+1)+xi$ and the right side is $(2-2x)-2yi$; equating parts gives $y+1=2-2x$ and $x=-2y$ (solved fully in the examples below).
Geometric equations: $|z-z_0|=r$ describes the circle of radius $r$ centered at $z_0$. If the equation contains $\bar z$, use $|\bar w|=|w|$ to rewrite it: for example$$|\bar z+i-1|=10 \iff |\overline{\bar z+i-1}|=10 \iff |z-i-1|=10,$$a circle of radius $10$ centered at $1+i$ (point $(1,1)$). Typical questions ask for the center, the radius, or the maximum/minimum of $|z|$ on the circle: $|z|_{\max}=|z_0|+r$, $|z|_{\min}=\bigl||z_0|-r\bigr|$.
Never expand $(1+i)^n$ term by term — square first:$$(1+i)^2=1+2i+i^2=2i,\qquad (1-i)^2=-2i.$$Chains that follow immediately:$$(1+i)^4=(2i)^2=-4,\qquad (1-i)^4=-4,\qquad (1+i)^8=16.$$Strategy for $z^n$: write $n=2m$ (or $n=2m+1$) and use $z^n=(z^2)^m$ or $z^n=(z^2)^m\cdot z$. Example: for $z=1+i$,$$z^6=(z^2)^3=(2i)^3=8i^3=-8i.$$The same idea works for other frequent bases: $(\sqrt3+i)^2=2+2\sqrt3\,i$, and for high powers switch to the trigonometric form (next section). Quick check with moduli: $|z^6|=|z|^6=(\sqrt2)^6=8$, consistent with $|-8i|=8$.
Every nonzero $z=a+bi$ can be written in trigonometric (polar) form:$$z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta,\qquad r=|z|=\sqrt{a^2+b^2},\ \ \tan\theta=\frac{b}{a},$$where $\theta$ (the argument) is chosen by the quadrant of the point $(a,b)$.
Multiplication and division act on moduli and angles:$$r_1\,\text{cis}\,\alpha\cdot r_2\,\text{cis}\,\beta=r_1r_2\,\text{cis}(\alpha+\beta),\qquad \frac{r_1\,\text{cis}\,\alpha}{r_2\,\text{cis}\,\beta}=\frac{r_1}{r_2}\,\text{cis}(\alpha-\beta).$$De Moivre's formula: $(r\,\text{cis}\,\theta)^n=r^n\,\text{cis}(n\theta)$.
Typical YÖS pattern: given $z_1=\text{cis}\,\alpha$, $z_2=\text{cis}\,\beta$, $z_3=\text{cis}\,\gamma$, an expression like $\dfrac{z_1z_3}{z_2}$ is just $\text{cis}(\alpha-\beta+\gamma)$ — add the angles in the numerator, subtract those in the denominator.
Anchor values: $\text{cis}\,0=1$, $\text{cis}\,90^\circ=i$, $\text{cis}\,180^\circ=-1$, $\text{cis}\,270^\circ=-i$, and $1+i=\sqrt2\,\text{cis}\,45^\circ$.
Problem A. Compute $i^{143}+i^{172}$.
Step 1. Reduce the exponents mod 4: $143=4\cdot 35+3$, so $i^{143}=i^3=-i$.Step 2. $172=4\cdot 43+0$, so $i^{172}=i^0=1$.Step 3. Add: $i^{143}+i^{172}=-i+1=1-i$.
Problem B. Compute $S=\sum_{k=1}^{73} i^k=i+i^2+\dots+i^{73}$.
Step 1. Any four consecutive powers sum to zero: $i+i^2+i^3+i^4=i-1-i+1=0$.Step 2. $73=4\cdot 18+1$: the first $72$ terms form $18$ complete zero blocks.Step 3. Only the last term survives: $S=i^{73}=i^{4\cdot 18+1}=i$.
Answer: $1-i$ and $i$.
Problem. Find $\text{Re}(z)$ if $z=\dfrac{(1+2i)(1-i)}{3-i}$.
Step 1. Expand the numerator:$$(1+2i)(1-i)=1-i+2i-2i^2=1+i+2=3+i.$$Step 2. Multiply by the conjugate of the denominator:$$z=\frac{3+i}{3-i}\cdot\frac{3+i}{3+i}=\frac{(3+i)^2}{3^2+1^2}=\frac{9+6i+i^2}{10}=\frac{8+6i}{10}.$$Step 3. Write in standard form: $z=\dfrac{4}{5}+\dfrac{3}{5}i$.
Answer: $\text{Re}(z)=\dfrac{4}{5}$ (and $\text{Im}(z)=\dfrac{3}{5}$).
The same three steps handle $\dfrac{i}{(1-i)(2+i)}$: expand $(1-i)(2+i)=3-i$ first, then $z=\dfrac{i}{3-i}=\dfrac{i(3+i)}{10}=\dfrac{-1+3i}{10}$, so $\text{Im}(z)=\dfrac{3}{10}$.
Problem. The complex number $z$ satisfies $i\bar z+1=2(1-z)$. Find $\text{Re}(z)$.
Step 1. Let $z=x+yi$, so $\bar z=x-yi$.Step 2. Left side: $i\bar z+1=i(x-yi)+1=ix-yi^2+1=(y+1)+xi$.Step 3. Right side: $2(1-z)=2-2x-2yi=(2-2x)+(-2y)i$.Step 4. Equate real and imaginary parts:$$y+1=2-2x,\qquad x=-2y.$$Step 5. Substitute $y=-\dfrac{x}{2}$ into the first equation: $-\dfrac{x}{2}+1=2-2x$, so $\dfrac{3x}{2}=1$ and $x=\dfrac{2}{3}$, $y=-\dfrac{1}{3}$.
Check: $z=\dfrac{2}{3}-\dfrac{1}{3}i$ gives $i\bar z+1=\dfrac{2}{3}+\dfrac{2}{3}i$ and $2(1-z)=\dfrac{2}{3}+\dfrac{2}{3}i$. Correct.
Answer: $\text{Re}(z)=\dfrac{2}{3}$.
Problem A. For $z=1+i$ compute $z^6$.
Step 1. Square first: $z^2=(1+i)^2=1+2i+i^2=2i$.Step 2. Then $z^6=(z^2)^3=(2i)^3=8i^3=-8i$.Step 3. Check via moduli: $|z|=\sqrt2$, so $|z^6|=(\sqrt2)^6=8=|-8i|$. Consistent.
Alternative via De Moivre: $z=\sqrt2\,\text{cis}\,45^\circ$, so $z^6=(\sqrt2)^6\,\text{cis}\,270^\circ=8(-i)=-8i$.
Problem B. Let $z_1=\text{cis}\,80^\circ$, $z_2=\text{cis}\,35^\circ$, $z_3=\text{cis}\,45^\circ$. Compute $w=\dfrac{z_1z_3}{z_2}$.
Step 1. Multiplication adds angles, division subtracts them: $w=\text{cis}(80^\circ+45^\circ-35^\circ)=\text{cis}\,90^\circ$.Step 2. $\text{cis}\,90^\circ=\cos 90^\circ+i\sin 90^\circ=i$.
Answers: $z^6=-8i$ and $w=i$.
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