H2 Math Complex Numbers: Study Guide

JC · H2 Mathematics · Study guide

Solve complex equations and polynomial replacements, then use modulus, argument and Argand vectors to recover exact values and square vertices.

Concepts · worked methods · graphs7 min read · practice linked below

Choose Cartesian form, factors or geometry

Use Cartesian form when conjugates or moduli remain after elimination. Use a quadratic equation when a variable and its square remain. For polynomial roots, first check that all coefficients are real before using the conjugate root theorem. On an Argand diagram, addition and subtraction represent vectors; multiplication by i rotates a vector anticlockwise through π/2.

Solve a quadratic with complex coefficients

Find a complex square root by comparing parts

Solve $z^2-(3+i)z+4+3i=0$ in Cartesian form.

\begin{aligned}\Delta&=(3+i)^2-4(4+3i),\\&=(8+6i)-(16+12i),\\&=-8-6i,\\z&=\frac{3+i\pm\sqrt\Delta}{2}.\end{aligned}

To find the square roots of $-8-6i$ , let $a+bi$ , with real $a,b$ , be one of them.

\begin{aligned}(a+bi)^2&=-8-6i,\\a^2-b^2&=-8,\\2ab&=-6.\end{aligned}

Also $a^2+b^2=|-8-6i|=10$ , since $|(a+bi)^2|=|a+bi|^2$ . Adding the two square equations:

\begin{aligned}2a^2&=2,\\a^2&=1,\quad b^2=9,\\ab&=-3.\end{aligned}

So the roots are $1-3i$ and $-1+3i$ . Use either with the ± in the quadratic formula:

\begin{aligned}z&=\frac{3+i\pm(1-3i)}2,\\z&=2-i\quad\text{or}\quad1+2i.\end{aligned}

Check their sum is $3+i$ and product is $4+3i$ . The coefficients are not all real, so the two roots need not be conjugates.

A non-real root of a real polynomial gives a factor

Find missing coefficients and all remaining roots

The polynomial $P(z)=z^4+Az^2+Bz+10$ , with real $A,B$ , has root $1+2i$ . Find $A,B$ and the other roots.

All coefficients are real, so $1-2i$ is also a root.

The factor from this pair is $Q(z)=[z-(1+2i)][z-(1-2i)]$ .

\begin{aligned}Q(z)&=(z-1)^2-(2i)^2,\\&=(z-1)^2+4,\\&=z^2-2z+5.\end{aligned}

Write $P(z)=(z^2-2z+5)(z^2+pz+q)$ . Compare the missing cubic coefficient and the constant first:

\begin{aligned}p-2&=0\implies p=2,\\5q&=10\implies q=2,\\A&=q-2p+5\\&=3,\\B&=-2q+5p\\&=6.\end{aligned}

The remaining quadratic gives

\begin{aligned}z^2+2z+2&=0,\\(z+1)^2&=-1,\\z&=-1\pm i.\end{aligned}

Thus $A=3,B=6$ ; the other roots are $1-2i,-1+i,-1-i$ . Expanding the factors checks the full polynomial.

Explanation: why can a quartic have at most two real roots?

If a real quartic has roots whose squares sum to a negative real number, all four roots cannot be real: real squares are non-negative. At least one root is non-real; its conjugate must also occur. At least two roots are therefore non-real, leaving at most two real roots, counted with multiplicity.

Hence roots: identify what replaces the original variable

Reversed coefficients suggest a reciprocal

The roots of $P(z)=z^3+z+10=0$ are $-2,1+2i,1-2i$ . Deduce the roots of $10w^3+w^2+1=0$ .

The constant term of $P$ is non-zero, so none of its roots is zero. Put $z=1/w$ :

\begin{aligned}P(1/w)&=\frac1{w^3}+\frac1w+10,\\w^3P(1/w)&=1+w^2+10w^3.\end{aligned}

Thus each required $w$ is the reciprocal of an original root:

\begin{aligned}w&=-\frac12,\\w&=\frac1{1+2i}\\&=\frac{1-2i}{5},\\w&=\frac1{1-2i}\\&=\frac{1+2i}{5}.\end{aligned}

These are all three roots. Check $w=-1/2$ : $10(-1/8)+1/4+1=0$ .

Variation: the new equation contains i

For $w^3-w+10i=0$ , note that

\begin{aligned}iP(iw)&=i(-iw^3+iw+10)\\&=w^3-w+10i.\end{aligned}

So $iw=z$ , or $w=-iz$ . The roots are

\begin{aligned}w&=2i,\quad2-i,\quad-2-i.\end{aligned}

Write the replacement equation before mapping roots; guessing the sign of the rotation can reverse the answer.

Eliminate the variable appearing without a conjugate

Only then write the remaining variable as x + iy

Solve $v+2w=7$ and $3v-w^*=5+5i$ .

Eliminate $v$ using three times the first equation minus the second:

\begin{aligned}3v+6w&=21,\\6w+w^*&=16-5i.\end{aligned}

Now let $w=x+iy$ , so $w^*=x-iy$ :

\begin{aligned}6(x+iy)+(x-iy)&=16-5i,\\7x+5iy&=16-5i.\end{aligned}

\begin{aligned}7x&=16,\\5y&=-5,\\x&=\frac{16}{7},\quad y=-1,\\w&=\frac{16}{7}-i,\\v&=7-2\left(\frac{16}{7}-i\right)\\&=\frac{17}{7}+2i.\end{aligned}

Check the second equation:

\begin{aligned}3v-w^*&=3\left(\frac{17}{7}+2i\right)\\&\quad-\left(\frac{16}{7}+i\right),\\&=5+5i.\end{aligned}

Conjugation changes the imaginary part's sign. It does not change the real part's sign.

If a quadratic remains, solve it as a quadratic

Use the modulus condition to choose between two assignments

Given $w+z^*=3-i$ , $wz^*=4$ , and $|w|<|z|$ , find $w,z$ .

Eliminate $z^*$ . There is no need to introduce four real unknowns.

\begin{aligned}z^*&=3-i-w,\\w(3-i-w)&=4,\\w^2-(3-i)w+4&=0.\end{aligned}

\begin{aligned}w&=\frac{3-i\pm\sqrt{-8-6i}}2,\\&=\frac{3-i\pm(1-3i)}2,\\w&=2-2i\quad\text{or}\quad1+i.\end{aligned}

If $w=2-2i$ , then $z^*=1+i$ , so $z=1-i$ . This gives $|w|=2\sqrt2>\sqrt2=|z|$ and must be rejected.

If $w=1+i$ , then

\begin{aligned}z^*&=3-i-(1+i)\\&=2-2i,\\z&=2+2i.\end{aligned}

Now $|w|=\sqrt2<2\sqrt2=|z|$ . Hence $w=1+i,z=2+2i$ . Their sum with $z^*$ and their product are $3-i$ and 4, respectively.

A modulus is a non-negative real number

Keep the sign condition before squaring

Solve $|z|+5w=0$ and $iz-2w=-2+3i$ .

The first equation shows $w=-|z|/5$ , a real number. Substitute it into the second before writing $z=x+iy$ :

\begin{aligned}iz+\frac25|z|&=-2+3i,\\-y+\frac25\sqrt{x^2+y^2}+ix&=-2+3i.\end{aligned}

Comparing imaginary parts gives $x=3$ . Comparing real parts gives

\begin{aligned}\frac25\sqrt{9+y^2}&=y-2,\\y&\geq2.\end{aligned}

The last condition is essential because the left side is non-negative. Now square:

\begin{aligned}4(9+y^2)&=25(y-2)^2,\\21y^2-100y+64&=0,\\(y-4)(21y-16)&=0,\\y&=4\quad\text{or}\quad\frac{16}{21}.\end{aligned}

Reject $16/21$ because it violates $y\geq2$ . Thus

\begin{aligned}z&=3+4i,\\|z|&=\sqrt{3^2+4^2}=5,\\w&=-1.\end{aligned}

Check: $|z|+5w=5-5=0$ and $i(3+4i)-2(-1)=-2+3i$ . Squaring alone would have admitted an invalid solution.

Use the quadrant when finding an argument

A quotient gives an exact trigonometric value

Let $z=-1+\sqrt3i$ and $w=\sqrt2-\sqrt2i$ . Find the principal argument of $z/w$ , and hence obtain an exact value of $\tan(\pi/12)$ .

\begin{aligned}|z|&=2,\quad\arg z=\frac{2\pi}{3},\\|w|&=2,\quad\arg w=-\frac\pi4,\\\arg(z/w)&=\frac{2\pi}{3}-\left(-\frac\pi4\right)\\&=\frac{11\pi}{12}.\end{aligned}

This lies in $(-\pi,\pi]$ , so no adjustment by $2\pi$ is needed. Rationalise the denominator:

\begin{aligned}\frac zw&=\frac{(-1+\sqrt3i)(1+i)}{2\sqrt2},\\&=-\frac{1+\sqrt3}{2\sqrt2}\\&\quad+i\frac{\sqrt3-1}{2\sqrt2},\\\tan\frac{11\pi}{12}&=-\frac{\sqrt3-1}{1+\sqrt3}\\&=\sqrt3-2.\end{aligned}

Since $11\pi/12=\pi-\pi/12$ , the required value is $\tan(\pi/12)=2-\sqrt3$ . The quotient is in quadrant II, so its tangent should be negative, agreeing with the Cartesian calculation. When adding or subtracting arguments, reduce the result to the required principal interval.

Variation: a sum of vectors has an unknown angle

Let $u=-\sqrt2+i\sqrt2$ , and let $v$ have modulus 3 and argument $\theta$ , with $0<\theta<\pi/4$ . Express $|u+v|^2$ in terms of $\theta$ .

Use the real and imaginary components, or the cosine rule on the parallelogram formed by $u$ and $v$ :

\begin{aligned}v&=3\cos\theta+3i\sin\theta,\\|u+v|^2&=(-\sqrt2+3\cos\theta)^2\\&\quad+(\sqrt2+3\sin\theta)^2,\\&=4+9(\cos^2\theta+\sin^2\theta)\\&\quad-6\sqrt2\cos\theta+6\sqrt2\sin\theta,\\&=13-12\cos(\theta+\pi/4).\end{aligned}

The angle between the vectors is $3\pi/4-\theta$ . The corresponding triangle's internal angle is its supplement, $\pi/4+\theta$ , which explains the sign in the cosine-rule version.

Opposite vertices of a square: rotate about the midpoint

Find the other vertices in the stated order

Points $A,C$ represent $a=2-i,c=-2+5i$ . Given that $ABCD$ is a square labelled anticlockwise, find $b,d$ .

The midpoint $M$ of the diagonal has complex number

\begin{aligned}m&=\frac{a+c}{2}\\&=2i,\\a-m&=2-3i.\end{aligned}

For anticlockwise ordering, rotate $\overrightarrow{MA}$ by $\pi/2$ to obtain $\overrightarrow{MB}$ :

\begin{aligned}b-m&=i(a-m),\\&=i(2-3i)\\&=3+2i,\\b&=3+4i,\\d&=2m-b\\&=4i-(3+4i)\\&=-3.\end{aligned}

Hence $b=3+4i,d=-3$ . Check the adjacent side vectors:

\begin{aligned}b-a&=1+5i,\\c-b&=-5+i\\&=i(1+5i).\end{aligned}

They have the same length, and the second is an anticlockwise quarter turn of the first. This checks both squareness and vertex order.

Quarter turns from MA:1

One anticlockwise quarter turn sends MA to MB: B represents 3 + 4i.

To rotate about $M$ , subtract $m$ , multiply the vector by $i$ , then add $m$ back. Multiplying the point's complex number directly by $i$ would rotate about the origin.

Before you finish

Are all coefficients real before invoking conjugate roots?
Did you return from z* to z at the end?
For a modulus equation, did you check every squared solution in the unsquared relation?
Is the argument in the correct quadrant and principal interval?
Does a geometric rotation use the correct centre and direction?

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