Complex numbers

Unit 1 Complex numbers

Key unit competence: Apply properties of operations of complex numbers in solving mathematical problems.

Key ideas

• The number $i$ satisfies $i^2=-1$. Powers repeat with period 4: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$.
• A complex number $z=a+bi$ has real part $\mathrm{Re}z=a$ and imaginary part $\mathrm{Im}z=b$ (a real number). It is drawn as the point $(a,b)$ on the Argand diagram.
• Conjugate $\bar{z}=a-bi$ (reflection in the real axis); modulus $|z|=\sqrt{a^2+b^2}$; $z\bar{z}=|z{|}^2$.
• Operations: add/subtract real and imaginary parts; multiply with $i^2=-1$; divide by multiplying top and bottom by the conjugate of the denominator.
• Square roots of $a+bi$: solve $x^2-y^2=a$, $2xy=b$, $x^2+y^2=\sqrt{a^2+b^2}$. Quadratics $a{z}^2+bz+c=0$ with $\Delta <0$ have roots $\frac{-b\pm i\sqrt{-\Delta }}{2a}$ (conjugate pair).
• Polar form $z=r(\cos \theta +i\sin \theta )=r\mathrm{cis}\theta =r{e}^{i\theta }$, $r=|z|$, $\theta =\arg z$ (principal value in $]-\pi ,\pi ]$). Moduli multiply and arguments add.
• De Moivre: $(\cos \theta +i\sin \theta {)}^n=\cos n\theta +i\sin n\theta$. $n$-th roots of $r\mathrm{cis}\theta$: $\sqrt[n]{r}\mathrm{cis}\frac{\theta +2k\pi }{n}$, $k=0,1,\dots ,n-1$; they form a regular $n$-gon.
• Loci: $|z-z_1|=R$ is a circle; $|z-z_1|=|z-z_2|$ is the perpendicular bisector; $\arg (z-z_1)=\theta$ is a half-line from $z_1$.
• Euler: $\cos x=\frac{e^{ix}+e^{-ix}}{2}$, $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ (used to linearise powers). $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos (x-\theta )$ where $\theta =\arg (a+bi)$.

Formulas to know

Common misconceptions

• "The imaginary part of $3-4i$ is $-4i$." No: it is the real number $-4$.
• "$\arg (-1-i)=\frac{\pi }{4}$." No: the point is in the third quadrant, so $\arg =-\frac{3\pi }{4}$. Always check the quadrant.
• "$z^3=1$ has only the root 1." No: there are three cube roots of unity.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Given $z_1=2+3i$ and $z_2=1-i$, find $z_1{z}_2$ and $\frac{z_1}{z_2}$.

   Show answer

   $z_1{z}_2=2-2i+3i+3=5+i$. $\frac{z_1}{z_2}=\frac{(2+3i)(1+i)}{2}=\frac{-1+5i}{2}=-\frac{1}{2}+\frac{5}{2}i$.

   Common mistake: Forgetting that $-3i^2=+3$.

2. Solve $\cos x+\sqrt{3}\sin x=1$.

   Show answer

   $|1+\sqrt{3}i|=2$, $\arg =\frac{\pi }{3}$: $2\cos \left(x-\frac{\pi }{3}\right)=1$, so $x-\frac{\pi }{3}=\pm \frac{\pi }{3}+2k\pi$: $x=\frac{2\pi }{3}+2k\pi$ or $x=2k\pi$.

   Common mistake: Giving only one family of solutions.

3. Solve $(1+i)z=3-i$.

   Show answer

   $z=\frac{3-i}{1+i}=\frac{(3-i)(1-i)}{2}=\frac{2-4i}{2}=1-2i$.

   Common mistake: Dividing real and imaginary parts separately.

4. Use De Moivre's theorem to express $\sin 3x$ in terms of $\sin x$.

   Show answer

   $\mathrm{Im}(\cos x+i\sin x{)}^3=3{\cos }^{2}x\sin x-{\sin }^{3}x=3(1-{\sin }^{2}x)\sin x-{\sin }^{3}x=3\sin x-4{\sin }^{3}x$.

   Common mistake: Sign error in $i^3=-i$.

5. Find the square roots of $-5+12i$.

   Show answer

   $x^2-y^2=-5$, $x^2+y^2=13$, $2xy=12$: $x^2=4$, $y^2=9$, $xy>0$. Roots: $\pm (2+3i)$.

   Common mistake: Ignoring the sign condition $xy>0$ and giving four roots.

6. Find the cube roots of $8i$ and plot them.

   Show answer

   $8i=8\mathrm{cis}\frac{\pi }{2}$. Roots $2\mathrm{cis}\left(\frac{\pi }{6}+\frac{2k\pi }{3}\right)$: $\sqrt{3}+i$, $-\sqrt{3}+i$, $-2i$. They are the vertices of an equilateral triangle.

   Common mistake: Dividing the argument by 3 but forgetting the $\frac{2k\pi }{3}$ terms.

7. Solve $z^2-(3+i)z+2+i=0$.

   Show answer

   $\Delta =(3+i{)}^2-4(2+i)=2i=(1+i{)}^2$. $z=\frac{3+i\pm (1+i)}{2}$: $z=2+i$ or $z=1$.

   Common mistake: Using $\sqrt{\Delta }$ as if $\Delta$ were a positive real number.

8. Write $z=1-\sqrt{3}i$ in polar and exponential form, then compute $z^6$.

   Show answer

   $r=2$, $\theta =-\frac{\pi }{3}$: $z=2\mathrm{cis}\left(-\frac{\pi }{3}\right)=2e^{-i\pi /3}$. $z^6=64\mathrm{cis}(-2\pi )=64$.

   Common mistake: Taking $\theta =\frac{\pi }{3}$ (wrong quadrant).

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Describe the set of points $z$ with $|z-1-i|=2$.	a. $11-10i$
   2. Compute $(1+i\sqrt{3}{)}^6$.	b. the circle with centre $(1,1)$ and radius 2
   3. Compute $(3+2i)(1-4i)$.	c. $2$ and $\frac{2\pi }{3}$
   4. Find the modulus and principal argument of $-1+i\sqrt{3}$.	d. $64$

   Show answer

   1 → b (the circle with centre $(1,1)$ and radius 2) · 2 → d ($64$) · 3 → a ($11-10i$) · 4 → c ($2$ and $\frac{2\pi }{3}$)

   • Step 1. $|z-(1+i)|=2$ is the distance from $1+i$.
     Step 2. All points at distance 2 from $(1,1)$ form a circle of radius 2.
   • Step 1. $1+i\sqrt{3}=2e^{i\pi /3}$.
     Step 2. De Moivre: $2^6{e}^{2\pi i}=64$.
   • Step 1. Expand: $3-12i+2i-8i^2$.
     Step 2. $i^2=-1$: $3+8-10i=11-10i$.
   • Step 1. $|z|=\sqrt{1+3}=2$.
     Step 2. The point is in quadrant II with $\tan \alpha =\sqrt{3}$: $\arg z=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\pm (1+2i)$  ·  $\pm (2+i)$  ·  $\pm 1\pm i$  ·  $a=-2$, $b=5$
   1. The square roots of $3+4i$ are .
   2. $1-2i$ is a root of $z^2+az+b=0$ with $a,b$ real. Then .
   3. The fourth roots of $-4$ are .
   Show answer

   1: $\pm (2+i)$ · 2: $a=-2$, $b=5$ · 3: $\pm 1\pm i$

   • Step 1. Let $(a+bi{)}^2=3+4i$: $a^2-b^2=3$, $2ab=4$.
     Step 2. $a=2$, $b=1$: $(2+i{)}^2=4+4i-1=3+4i$. The roots are $\pm (2+i)$.
   • Step 1. Real coefficients: the other root is $1+2i$.
     Step 2. Sum $=2=-a$, so $a=-2$. Product $=1+4=5=b$.
   • Step 1. $-4=4e^{i\pi }$; roots $\sqrt{2}{e}^{i(\pi /4+k\pi /2)}$, $k=0,1,2,3$.
     Step 2. $\sqrt{2}{e}^{i\pi /4}=1+i$, and the others are $-1+i$, $-1-i$, $1-i$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Simplify $e^{i\pi /3}+e^{-i\pi /3}$.	a. $18$
   2. Find the square roots of $-5+12i$.	b. $1$
   3. $|z|=3$ and $|w|=2$. Find $|z^2\bar{w}|$.	c. $\pm (2+3i)$
   4. $z=e^{i\theta }$. Then $z^3+\frac{1}{z^3}=$	d. $2\cos 3\theta$

   Show answer

   1 → b ($1$) · 2 → c ($\pm (2+3i)$) · 3 → a ($18$) · 4 → d ($2\cos 3\theta$)

   • Step 1. $e^{i\theta }+e^{-i\theta }=2\cos \theta$.
     Step 2. $2\cos \frac{\pi }{3}=2\times \frac{1}{2}=1$.
   • Step 1. Let $(a+bi{)}^2=-5+12i$: $a^2-b^2=-5$ and $2ab=12$.
     Step 2. $a=2$, $b=3$ works: $(2+3i{)}^2=4+12i-9=-5+12i$.
   • Step 1. $|z^2\bar{w}|=|z{|}^2|\bar{w}|$.
     Step 2. $|\bar{w}|=|w|=2$, so $9\times 2=18$.
   • Step 1. $z^3=e^{3i\theta }$ and $z^{-3}=e^{-3i\theta }$.
     Step 2. Their sum is $2\cos 3\theta$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $0$  ·  $z=\pm (1-i)$  ·  $z=2\pm \sqrt{3}i$  ·  $z=2\pm 3i$
   1. The sum of the five fifth roots of unity is .
   2. Solve $z^2=-2i$. →
   3. Solve $z^2-4z+13=0$. →
   Show answer

   1: $0$ · 2: $z=\pm (1-i)$ · 3: $z=2\pm 3i$

   • Step 1. They are the roots of $z^5-1=0$.
     Step 2. The sum of the roots is $-\frac{\text{coefficient of }{z}^4}{1}=0$.
   • Step 1. Let $(a+bi{)}^2=-2i$: $a^2-b^2=0$, $2ab=-2$.
     Step 2. $a=1$, $b=-1$: $(1-i{)}^2=1-2i-1=-2i$.
   • Step 1. $\Delta =16-52=-36=(6i{)}^2$.
     Step 2. $z=\frac{4\pm 6i}{2}=2\pm 3i$.

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