Advanced Higher Maths
Complex Numbers

(a)  To find a complex conjugate, we just change the sign of the imaginary part. So \(\overline{z}=2+\sqrt{3}\,i\small.\)

(b)  To find \(\raise 0.1pt{z\overline{z}}\small,\) we just break the brackets, using the fact that \(i^2=-1\small.\)

$$ \begin{eqnarray} z\overline{z} &=& \left(2-\sqrt{3}\,i\right)\left(2+\sqrt{3}\,i\right)\\[6pt] &=& 2^2-3\,i^2 \\[6pt] &=& 4+3 \\[6pt] &=& 7 \end{eqnarray} $$

Note: For any complex number \(z\small,\) it is always the case that \(\raise 0.1pt{z\overline{z}}\) is a real number, for much the same reason as why a difference of two squares has no 'middle term'.

(a)  The complex conjugate \(\overline{z_2}=k+12i\small.\) So:

$$ \begin{eqnarray} z_{1}\overline{z_2} &=& (3+4i)(k+12i)\\[6pt] &=& 3k+36i+4ki+48i^2 \\[6pt] &=& (3k-48)+(4k+36)i \end{eqnarray} $$

(b)  \(z_{1}\overline{z_2}\in\mathbb R\) when its imaginary part equals zero.

$$ \begin{gather} 4k+36=0\\[6pt] k=-9 \end{gather} $$

(a)  We will need to perform the division in order to identify the real and imaginary parts of \(z\small.\)

To divide by a complex number we multiply top and bottom by the complex conjugate of the divisor.

$$ \begin{eqnarray} z &=& \frac{3-i}{2+ni} \\[12pt] &=& \frac{3-i}{2+ni}\normalsize\times\frac{2-ni}{2-ni} \\[12pt] &=& \frac{(3-i)(2-ni)}{(2+ni)(2-ni)} \\[12pt] &=& \frac{6-2i-3ni-n}{4+n^2} \\[12pt] &=& \frac{6-n}{4+n^2} + \left(\frac{-2-3n}{4+n^2}\right)i \end{eqnarray} $$

\(z\in\mathbb R\) so \(\text{Im}(z)=0\small,\) so the numerator of the imaginary part equals zero, and we solve as follows:

$$ \begin{gather} -2-3n=0\\[8pt] -3n=2\\[6pt] n=-\small\frac{2}{3} \end{gather} $$

(b)  Now we can substitute the value of \(n\) into \(\text{Re}(z)\small.\)

$$ \begin{eqnarray} z &=& \frac{6-n}{4+n^2}\\[10pt] &=& \frac{6+\large{{}^{2}\!/\!_{3}}}{4+\left(\large{{}^{2}\!/\!_{3}}\right)^2} \\[12pt] &=& \frac{\large{{}^{20}\!/\!_{3}}}{4+\large{{}^{4}\!/\!_{9}}} \\[12pt] &=& \frac{20}{3}\div{\frac{40}{9}} \\[12pt] &=& \frac{20}{3}\times{\frac{9}{40}} \\[12pt] &=& \frac{3}{2} \end{eqnarray} $$

We use the quadratic formula with \(a=1\), \(b=-4\) and \(c=5\small.\)

Once we find that we have a negative discrimant, we just continue in terms of \(i\small.\)

$$ \begin{eqnarray} x &=& \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[10pt] &=& \frac{4\pm\sqrt{(-4)^2-4(1)(5)}}{2}\\[10pt] &=& \frac{4\pm\sqrt{16-20}}{2}\\[10pt] &=& \frac{4\pm\sqrt{-4}}{2}\\[10pt] &=& \frac{4\pm\sqrt{4}\sqrt{-1}}{2}\\[10pt] &=& \frac{4\pm 2i}{2}\\[10pt] &=& 2\pm i \end{eqnarray} $$

We start by letting \(z=x+yi\small,\) say.

So the complex conjugate \(\overline{z}=x-yi\small.\)

The equation then becomes:

$$ \begin{gather} (x+yi) +2i(x-yi) = 8+7i\\[6pt] x+yi+2xi-2yi^2 = 8+7i\\[6pt] x+yi+2xi+2y = 8+7i\\[6pt] (x+2y)+(yi+2xi) = 8+7i\\[6pt] (x+2y)+(y+2x)i = 8+7i \end{gather} $$

Equating the real parts, we obtain this equation:

$$x+2y=8$$

Equating the imaginary parts, we obtain a second equation:

$$y+2x=7$$

All that remains is to solve these two equations simultaneously to find \(x\) and \(y\small.\)

One way of doing this is to rearrange the second equation to \(y=7-2x\) and substitute into the first equation:

$$ \begin{gather} x+2(7-2x) = 8\\[6pt] x+14-4x = 8\\[6pt] -3x = 8-14\\[6pt] -3x = -6\\[6pt] x = 2 \end{gather} $$

So \(y=7-2(2)=3\) which gives us \(z = 2+3i\small.\)

It is important to remember that there are two square roots of any non-zero complex number.

We start by writing \(\sqrt{3-4i\small\,\normalsize}\) as \(a+bi\small,\) where \(\raise 0.1pt{a\small,\normalsize\,b\in\mathbb R}\small.\)

We then square both sides, to obtain an expression for \(3-4i\) in terms of \(a\) and \(b\):

$$ \begin{eqnarray} 3-4i &=& (a+bi)^2\\[6pt] &=& a^2+2abi-b^2\\[6pt] &=& (a^2-b^2)+2abi \end{eqnarray} $$

Now we start working towards finding the values of \(a\) and \(b\small.\) The first step in this process is to equate the imaginary part of the equation, to obtain an expression for \(b\) in terms of \(a\):

$$ \begin{eqnarray} -4 &=& 2ab\\[6pt] b &=& -\!\frac{2}{a} \end{eqnarray} $$

Then we substitute this into the real part of the equation:

$$ \begin{gather} 3 = a^2-b^2\\[6pt] 3 = a^2-\left(-\frac{2}{a}\right)^2\\[6pt] 3 = a^2-\frac{4}{a^2}\\[8pt] 3a^2 = a^4-4\\[8pt] a^4-3a^2-4 = 0\\[8pt] (a^2+1)(a^2-4) = 0\\[8pt] a^2+1=0\:\:\small\textsf{or}\normalsize\:\:a^2-4=0\\[8pt] \end{gather} $$

\(a^2+1=0\) has no real solutions, but \(a^2-4=0\) gives us \(a=\pm 2\small.\)

Finally, we find the corresponding \(b\)-values:

\(a=-2\implies b=-\left(\large\frac{2}{-2}\normalsize\!\right)=1\small,\) giving a root of \(-2+i\small.\)

\(a=2\implies b=-\large\frac{2}{2}\normalsize=-1\small,\) giving the other root: \(2-i\small.\)

Note: The two square roots of a non-zero complex number have the same modulus but are diametrically opposed across the origin. In polar form, this means that their arguments differ by exactly \(\pi\) radians \((180^{\circ})\small.\)

As \(z\) is a root, \(\overline{z}=1-2i\) is another root. Complex roots always exist in conjugate pairs.

Therefore \(\bigl(z-(1\!+\!2i)\bigr)\) and \(\bigl(z-(1\!-\!2i)\bigr)\) are factors of \(z^3-5z^2+11z-15\small.\)

We now multiply these two linear factors to obtain a quadratic factor, using the regrouping method below:

$$ \begin{eqnarray} && \!\!\!\!\!\!\bigl(z-(1\!+\!2i)\bigr)\bigl(z-(1\!-\!2i)\bigr)\\[8pt] &=& \bigl(z-1-2i\bigr)\bigl(z-1+2i\bigr)\\[8pt] &=& \bigl((z-1)-2i\bigr)\bigl((z-1)+2i\bigr)\\[8pt] &=& (z-1)^2-4i^2 \\[8pt] &=& z^2-2z+1+4 \\[8pt] &=& z^2-2z+5 \\[8pt] \end{eqnarray} $$

Now we use polynomial long division to find the third factor.

So the final factor is \((z-3)\) and now we can list all three roots:

\(z=1+2i\small,\) \(\,z=1-2i\small,\) \(\,z=3\small.\)

The complex conjugate \(\overline{z}=1+\sqrt{3}\tiny\,\normalsize i\,\) is another root.

So \(\bigl(z-(1\!-\!\sqrt{3}\,)\bigr)\) and \(\bigl(z-(1\!+\!\sqrt{3}\,i)\bigr)\) are factors of \(z^4+3z^2+2z+12\small.\)

We now multiply these two linear factors to obtain a quadratic factor:

$$ \begin{eqnarray} && \!\!\!\!\!\!\biggl(z-\left(1\!-\!\sqrt{3}\tiny\,\normalsize i\right)\biggr)\biggl(z-\left(1\!+\!\sqrt{3}\tiny\,\normalsize i\right)\biggr)\\[6pt] &=& \biggl(z-1+\sqrt{3}\tiny\,\normalsize i\biggr)\biggl(z-1-\sqrt{3}\tiny\,\normalsize i\biggr)\\[6pt] &=& \biggl(\left(z-1\right)+\sqrt{3}\tiny\,\normalsize i\biggr)\biggl(\left(z-1\right)-\sqrt{3}\tiny\,\normalsize i\biggr)\\[6pt] &=& (z-1)^2 -3i^2 \\[6pt] &=& z^2-2z+1+3 \\[6pt] &=& z^2-2z+4 \\[6pt] \end{eqnarray} $$

Next, we use polynomial long division, taking care not to forget the missing \(z^3\) term:

Next, we use the quadratic formula to find the roots of \(z^2+2z+3=0\small.\)

$$ \begin{eqnarray} z&=& \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\[6pt] &=& \frac{-2\pm\sqrt{2^{2}-4\!\times\!1\!\times\!3}}{2}\\[6pt] &=& \frac{-2\pm\sqrt{-8}}{2}\\[6pt] &=& \frac{-2\pm\sqrt{8}\,i}{2}\\[6pt] &=& \frac{-2\pm2\sqrt{2}\,i}{2}\\[6pt] &=& -\!1\pm\sqrt{2}\tiny\,\normalsize i \end{eqnarray} $$

So now we can list all four roots:
\(z=1\pm\sqrt{3}\,i\,\) and \(\,z=-1\pm\sqrt{2}\,i\small.\)

(a)  Cartesian form: \(z=k\sqrt{3}-k\tiny\,\normalsize i\small.\) This would only be worth one mark.

(b)  Polar form. This would typically be worth three marks: one for the modulus \(\vert z\vert\small,\) one for the argument \(\text{arg}(z)\) and one for correctly combining them into polar form.

For the modulus, we use the distance formula (Pythagoras):

$$ \begin{eqnarray} \vert z\vert\ &=& \sqrt{\left(k\sqrt{3}\right)^2+(-k)^2}\\[6pt] &=& \sqrt{(3k^2+k^{2})}\\[6pt] &=& \sqrt{4k^{2}}\\[8pt] &=& \ 2k\\[12pt] \end{eqnarray} $$

To find the principal argument, in the interval \(-\pi\lt\text{arg}(z)\leqslant\pi\):

$$ \begin{eqnarray} \text{arg}(z) &=& \text{tan}^{-1}\small\left(\frac{-k}{k\sqrt{3}}\right)\normalsize \\[6pt] &=& \text{tan}^{-1}\small\left(\frac{-1}{\sqrt{3}}\right)\normalsize \\[6pt] &=& -\!\frac{\pi}{6} \\[6pt] \end{eqnarray} $$

So the polar form is:

$$ z=2k\,\biggl(\text{cos}\left(-\small\frac{\pi}{6}\normalsize\right)+i\,\text{sin}\left(-\small\frac{\pi}{6}\normalsize\right)\biggr) $$

Note 1: Time permitting, it may be a good idea in your exam to check that your polar form is correct by evaluating each trig expression and checking that it simplifies back into the Cartesian form – but be careful to score out your check, or otherwise make it very clear what your final answer is, in case the marker isn't sure that you knew where to stop!

Note 2: You may be tempted to use your knowledge from the Functions topic that cosine is even and sine is odd to simplify the polar form to the following:

$$ z=2k\,\biggl(\text{cos}\,\small\frac{\pi}{6}\normalsize-i\,\text{sin}\,\small\frac{\pi}{6}\normalsize\biggr) $$

However, this isn't technically in polar form any more, because of the minus sign before the imaginary term. So, if the polar form has a negative argument, just leave it, despite the clumsy appearance of the nested brackets.

(a)  To multiply two numbers in polar form we multiply their moduli and add their arguments, if necessary bringing the argument back into the principal interval \(-\pi\lt\text{arg}(z)\leqslant\pi\) by adding or subtracting multiples of \(2\pi\) as required. We will need to subtract \(2\pi\) in this case.

$$ \begin{eqnarray} zw &=& 6\,\biggl(\text{cos}\small\left(\frac{\pi}{4}\!+\!\frac{5\pi}{6}\right)\normalsize+i\ \text{sin}\small\left(\frac{\pi}{4}\!+\!\frac{5\pi}{6}\right)\normalsize\biggr)\\[10pt] &=& 6\,\biggl(\text{cos}\small\left(\frac{3\pi\!+\!10\pi}{12}\right)\normalsize+i\ \text{sin}\small\left(\frac{3\pi\!+\!10\pi}{12}\right)\normalsize\biggr)\\[10pt] &=& 6\,\biggl(\text{cos}\,\small\frac{13\pi}{12}\normalsize+i\ \text{sin}\,\small\frac{13\pi}{12}\normalsize\biggr)\\[10pt] &=& 6\,\biggl(\text{cos}\small\left(\!\frac{13\pi}{12}\normalsize\!-\!2\pi\small\right)\normalsize+i\ \text{sin}\small\left(\!\frac{13\pi}{12}\normalsize\!-\!2\pi\small\right)\normalsize\biggr)\\[10pt] &=& 6\,\biggl(\text{cos}\small\left(\!\frac{13\pi\!-\!24\pi}{12}\!\right)\normalsize\!+i\ \text{sin}\small\left(\!\frac{13\pi\!-\!24\pi}{12}\right)\normalsize\biggr)\\[10pt] &=& 6\,\biggl(\text{cos}\small\left(-\frac{11\pi}{12}\right)\normalsize\!+i\ \text{sin}\small\left(-\frac{11\pi}{12}\right)\normalsize\biggr) \end{eqnarray} $$

(b)  To divide two numbers in polar form we divide their moduli and subtract their arguments, if necessary bringing the argument back into the principal interval \(-\pi\lt\text{arg}(z)\leqslant\pi\) by adding or subtracting multiples of \(2\pi\small.\) That won't be necessary in this case.

$$ \begin{eqnarray} \small\frac{z}{w}\normalsize &=& \small\frac{2}{3}\,\normalsize\biggl(\text{cos}\small\left(\frac{\pi}{4}\!-\!\frac{5\pi}{6}\right)\normalsize+i\ \text{sin}\small\left(\frac{\pi}{4}\!-\!\frac{5\pi}{6}\right)\normalsize\biggr)\\[10pt] &=& \small\frac{2}{3}\,\normalsize\biggl(\text{cos}\small\left(\frac{3\pi\!-\!10\pi}{12}\right)\normalsize+i\ \text{sin}\small\left(\frac{3\pi\!-\!10\pi}{12}\right)\normalsize\biggr)\\[10pt] &=& \small\frac{2}{3}\,\normalsize\biggl(\text{cos}\small\left(-\frac{7\pi}{12}\right)\normalsize\!+i\ \text{sin}\small\left(-\frac{7\pi}{12}\right)\normalsize\biggr) \end{eqnarray} $$

This solution will require de Moivre's theorem, so we first express \(\raise 0.1pt{z}\) in polar form.

$$ \begin{eqnarray} \vert z\vert\ &=& \left|\,-1-i\,\right|\\[6pt] &=& \sqrt{(-1)^2+(-1)^{2}}\\[6pt] &=& \sqrt{2} \end{eqnarray} $$

To find the principal argument, in the interval \(-\pi\lt\text{arg}(z)\leqslant\pi\small,\) we first find the related acute angle, and then adjust so that it is in the correct quadrant.

In this case, the argument is in the third quadrant (bottom left), so one way of doing this is to use \(\pi\) plus the acute angle and then subtract \(2\pi\) to bring it into the principal interval:

$$ \begin{eqnarray} \text{arg}(z) &=& \pi+\text{tan}^{-1}\left|\small\frac{-1}{-1}\normalsize\right|-2\pi \\[6pt] &=& \pi+\small\frac{\pi}{4}\normalsize-2\pi \\[6pt] &=& \small\frac{\pi}{4}\normalsize-\pi \\[6pt] &=& -\!\small\frac{3\pi}{4}\normalsize \\[6pt] \end{eqnarray} $$

\(\therefore\:z=\sqrt{2}\,\biggl(\text{cos}\left(-\large\frac{3\pi}{4}\normalsize\right)+i\ \text{sin}\left(-\large\frac{3\pi}{4}\normalsize\right)\biggr)\)

Raising \(\raise 0.1pt{z}\) to the power of \(10\) requires de Moivre's theorem:

$$ \bigl[r\,(\text{cos}\,\theta+i\,\text{sin}\,\theta)\bigr]^n = r^{n}\left(\text{cos}\,n\theta+i\,\text{sin}\,n\theta\right) $$

Substituting \(r=\sqrt{2}\small,\,\) \(\theta=-\large\frac{3\pi}{4}\) and \(n=10\) gives:

$$ \begin{eqnarray} z^{10} &=& \left(\sqrt{2}\right)^{10}\,\biggl(\text{cos}\left(-\small\frac{30\pi}{4}\normalsize\right)+i\ \text{sin}\left(-\small\frac{30\pi}{4}\normalsize\right)\biggr)\\[6pt] &=& 2^5\,\biggl(\text{cos}\left(-\small\frac{15\pi}{2}\normalsize\right)+i\ \text{sin}\left(-\small\frac{15\pi}{2}\normalsize\right)\biggr) \end{eqnarray} $$

The argument \(-\large\frac{15\pi}{2}\) is outside the principal interval, so we adjust it:

$$ \small-\frac{15\pi}{2}\normalsize+4(2\pi) = \small\frac{\pi}{2} $$

Now we can write our final answer:

$$ z^{10}=32\left(\text{cos}\small\,\frac{\pi}{2}\normalsize\!+i\,\text{sin}\small\,\frac{\pi}{2}\normalsize\right) $$

Let \(z^4=-1+i\small.\,\) We will find all values of \(z\small.\)

First we express \(z^4\) in polar form:

$$ \begin{eqnarray} \left|z^4\right| &=& \sqrt{(-1)^2+1^{2}}\\[6pt] &=& \sqrt{2} \end{eqnarray} $$

The argument is in the second quadrant, so:

$$ \begin{eqnarray} \text{arg}(z^4) &=& \pi-\text{tan}^{-1}\left|\small\frac{1}{-1}\right| \\[6pt] &=& \pi-\small\frac{\pi}{4} \\[6pt] &=& \small\frac{3\pi}{4} \\[6pt] \end{eqnarray} $$

\(\therefore\:z^4=\sqrt{2}\,\left(\text{cos}\small\,\large\frac{3\pi}{4}\normalsize+i\ \text{sin}\small\,\large\frac{3\pi}{4}\normalsize\right)\)

The fourth roots are as follows:

$$ z = 2^{\large\frac{1}{8}\normalsize}\biggl(\text{cos}\left(\small\frac{3\pi}{4}\normalsize\!+\!2k\pi\right)\normalsize+i\ \text{sin}\left(\small\frac{3\pi}{4}\normalsize\!+\!2k\pi\right)\biggr)^{\large\frac{1}{4}\normalsize} $$

where \(k=0\small,\normalsize\ 1\small,\normalsize\ 2\small,\normalsize\ 3\small.\)

Now we apply de Moivre's theorem:

$$ z = 2^{\large\frac{1}{8}\normalsize}\biggl(\text{cos}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{k\pi}{2}\normalsize\right)\normalsize+i\ \text{sin}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{k\pi}{2}\normalsize\right)\biggr) $$

where \(k=0\small,\normalsize\ 1\small,\normalsize\ 2\small,\normalsize\ 3\small.\)

To obtain separate expressions for each of these roots, we substitute each value of \(\raise 0.1pt{k}\) in turn, if necessary adjusting the argument in the usual way to ensure that it is within the principal interval.

When \(k=0\):

$$ \begin{eqnarray} z_1 &=& 2^{\large\frac{1}{8}\normalsize}\biggl(\text{cos}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{0\pi}{2}\normalsize\right)+i\ \text{sin}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{0\pi}{2}\normalsize\right)\biggr) \\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\left(\text{cos}\,\small\frac{3\pi}{16}\normalsize+i\ \text{sin}\,\small\frac{3\pi}{16}\normalsize\right) \end{eqnarray} $$

When \(k=1\):

$$ \begin{eqnarray} z_2 &=& 2^{\large\frac{1}{8}\normalsize}\biggl(\text{cos}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{1\pi}{2}\normalsize\right)+i\ \text{sin}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{1\pi}{2}\normalsize\right)\biggr) \\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\left(\text{cos}\,\small\frac{11\pi}{16}\normalsize+i\ \text{sin}\,\small\frac{11\pi}{16}\normalsize\right) \end{eqnarray} $$

When \(k=2\):

$$ \begin{eqnarray} z_3 &=& 2^{\large\frac{1}{8}\normalsize}\biggl(\text{cos}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{2\pi}{2}\normalsize\right)+i\ \text{sin}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{2\pi}{2}\normalsize\right)\biggr) \\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\left(\text{cos}\,\small\frac{19\pi}{16}\normalsize+i\ \text{sin}\,\small\frac{19\pi}{16}\normalsize\right)\\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\,\biggl(\text{cos}\left(\small\frac{19\pi\!-\!32\pi}{16}\normalsize\right)+i\ \text{sin}\left(\small\frac{19\pi\!-\!32\pi}{16}\right)\normalsize\biggr)\\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\,\biggl(\text{cos}\,\small\left(\frac{-13\pi}{16}\right)\normalsize+i\ \text{sin}\,\small\left(\frac{-13\pi}{16}\right)\normalsize\biggr)\\[6pt] \end{eqnarray} $$

When \(k=3\):

$$ \begin{eqnarray} z_4 &=& 2^{\large\frac{1}{8}\normalsize}\biggl(\text{cos}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{3\pi}{2}\normalsize\right)+i\ \text{sin}\left(\small\frac{3\pi}{16}\normalsize\!+\!\small\frac{3\pi}{2}\normalsize\right)\biggr) \\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\left(\text{cos}\,\small\frac{27\pi}{16}\normalsize+i\ \text{sin}\,\small\frac{27\pi}{16}\normalsize\right)\\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\,\biggl(\text{cos}\left(\small\frac{27\pi\!-\!32\pi}{16}\normalsize\right)+i\ \text{sin}\left(\small\frac{27\pi\!-\!32\pi}{16}\normalsize\right)\biggr)\\[6pt] &=& \sqrt[\uproot{3}\scriptstyle 8]{2}\,\biggl(\text{cos}\,\small\left(\frac{-5\pi}{16}\right)\normalsize+i\ \text{sin}\,\small\left(\frac{-5\pi}{16}\right)\normalsize\biggr)\\[6pt] \end{eqnarray} $$

The complex conjugate \(\overline{z_2}=6-2i\,\small.\)

$$ \begin{eqnarray} \therefore\:z_{1}\overline{z_2} &=& (5+3i)(6-2i)\\[6pt] &=& 30-10i+18i-6i^2\\[6pt] &=& 30+8i+6\\[6pt] &=& 36+8i \end{eqnarray} $$

(a)  For the modulus, we use the distance formula (Pythagoras):

$$ \begin{eqnarray} \vert z\vert\ &=& \sqrt{1^2+\left(\sqrt{3}\right)^2}\\[6pt] &=& \sqrt{1+3}\\[6pt] &=& \sqrt{4}\\[8pt] &=& \ 2\\[12pt] \end{eqnarray} $$

To find the principal argument, in the interval \(-\pi\lt\text{arg}(z)\leqslant\pi\):

$$ \begin{eqnarray} \text{arg}(z) &=& \text{tan}^{-1}\small\left(\frac{\sqrt{3}}{1}\right)\normalsize \\[6pt] &=& \text{tan}^{-1}\,\sqrt{3}\normalsize \\[6pt] &=& \frac{\pi}{3} \\[6pt] \end{eqnarray} $$

So the polar form is:

$$ z=2\left(\text{cos}\,\small\frac{\pi}{3}\normalsize +i\,\text{sin}\,\small\frac{\pi}{3}\normalsize\right) $$

(b)  Using de Moivre's theorem to find \(z^3\):

$$ \bigl[r\,(\text{cos}\,\theta+i\,\text{sin}\,\theta)\bigr]^n = r^{n}\left(\text{cos}\,n\theta+i\,\text{sin}\,n\theta\right) $$

Substituting \(r=2\small,\,\) \(\theta=\large\frac{\pi}{3}\) and \(n=3\) gives:

$$ \begin{eqnarray} z^3 &=& 2^3\small\,\normalsize\left(\text{cos}\,\small\frac{3\pi}{3}\normalsize\!+i\,\text{sin}\,\small\frac{3\pi}{3}\right)\\[6pt] &=& 8\left(\text{cos}\,\pi+i\,\text{sin}\,\pi\right)\\[6pt] &=& 8\left(-1+i(0)\right)\\[6pt] &=& \!-8 \in\mathbb R \end{eqnarray} $$

Note: The (b) part is one of those rare instances in which an 'or otherwise' method would be fine. We could expand the Cartesian form either using basic complex arithmetic or with the Binomial Theorem. Either of these alternative methods would have received full credit. Why not try them yourself as an exercise?

(a)  For the modulus, we use the distance formula (Pythagoras):

$$ \begin{eqnarray} \vert z\vert\ &=& \sqrt{1^2+1^2}\\[6pt] &=& \sqrt{2}\\[12pt] \end{eqnarray} $$

To find the principal argument, in the interval \(-\pi\lt\text{arg}(z)\leqslant\pi\small,\):

$$ \begin{eqnarray} \text{arg}(z) &=& \text{tan}^{-1}\small\left(\frac{1}{1}\right)\normalsize \\[6pt] &=& \frac{\pi}{4} \\[6pt] \end{eqnarray} $$

So the polar form is:

$$ z=\sqrt{2}\left(cos\,\small\frac{\pi}{4}\normalsize +i\,sin\,\small\frac{\pi}{4}\normalsize\right) $$

(b)  de Moivre's theorem states:

$$ \bigl[r\,(\text{cos}\,\theta+i\,\text{sin}\,\theta)\bigr]^n = r^{n}\left(\text{cos}\,n\theta+i\,\text{sin}\,n\theta\right) $$

Substituting \(r=\sqrt{2}\small,\) \(\theta=\large\frac{\pi}{4}\) and \(n=8\) gives:

$$ \begin{eqnarray} z^8 &=& \left(\sqrt{2}\right)^8\left(\text{cos}\,\small\frac{8\pi}{4}\normalsize+i\ \text{sin}\,\small\frac{8\pi}{4}\right)\\[6pt] &=& 16\small\,\normalsize\bigl(\text{cos}\,2\pi\!+i\,\text{sin}\,2\pi\bigr)\\[6pt] &=& 16\small\,\normalsize\bigl(1\!+i\,(0)\bigr)\\[6pt] &=& 16 \end{eqnarray} $$

The complex conjugate \(\overline{z}=x-iy\small.\)

So the equation is:

$$ \begin{gather} (x+iy)^2+20(x-iy)-156=0\\[6pt] x^2+2ixy+i^{2}y^{2}+20x-20iy-156=0\\[6pt] x^2+2ixy-y^2+20x-20iy-156=0\\[6pt] \end{gather} $$

Equating the imaginary parts:

$$ \begin{gather} 2xy-20y=0\\[6pt] y(2x-20)=0\\[6pt] 2x-20=0\\[6pt] x=10\\[6pt] \end{gather} $$

Note that we can only divide by \(y\) above as we know that \(y\neq 0\small.\)

Now, equating the real parts:

$$ \begin{gather} x^2-y^2+20x-156=0\\[6pt] 10^2-y^2+20(10)-156=0\\[6pt] 100-y^2+200-156=0\\[6pt] 144-y^2=0\\[6pt] y^2=144\\[6pt] y=\pm 12\\[6pt] \end{gather} $$

So \(z=10\pm 12i\,\small.\)

To divide complex numbers we multiply top and bottom by the complex conjugate of the denominator.

$$ \begin{eqnarray} \frac{z}{w} &=& \frac{11+10i}{3-2i} \\[12pt] &=& \frac{11+10i}{3-2i}\times\frac{3+2i}{3+2i} \\[12pt] &=& \frac{33+22i+30i+20i^2}{3^2-4i^2} \\[12pt] &=& \frac{33+52i-20}{9+4} \\[12pt] &=& \frac{13+52i}{13} \\[12pt] &=& 1+4i \\[12pt] \end{eqnarray} $$