The complex conjugate \(\overline{z}=2+\sqrt{3}\tiny\,\normalsize i\small.\)

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

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

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

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

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

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

$$ \begin{eqnarray} z=\small\frac{3-i}{2+ni}\normalsize &=& \small\frac{3-i}{2+ni}\normalsize\times\small\frac{2-ni}{2-ni}\normalsize\\[9pt] &=& \small\frac{(3-i)(2-ni)}{(2+ni)(2-ni)} \\[9pt] &=& \small\frac{(6-2i-3ni-n)}{(2^2+n^2)} \\[9pt] &=& \small\frac{6-n}{4+n^2}\normalsize + \small\frac{-2-3n}{4+n^2}\normalsize i \end{eqnarray} $$

\(z\in\mathbb R\) so \(\small\textsf{Im}\normalsize(z)=0\) and we solve for \(n\small.\)

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

Now we can subtitute into \(\small\textsf{Re}\normalsize(z)\small.\)

$$ \begin{eqnarray} z &=& \small\frac{6-n}{4+n^2}\\[6pt] &=& \small\frac{6+\frac{2}{3}}{4+\left(\frac{2}{3}\right)^2} \\[8pt] &=& \small\frac{\frac{20}{3}}{4\frac{4}{9}} \\[8pt] &=& \small\frac{\frac{60}{9}}{\frac{40}{9}} \\[8pt] &=& \small\frac{60}{40} \\[8pt] &=& \small\frac{3}{2} \end{eqnarray} $$

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

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

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:

$$x+2y=8$$

$$y+2x=7$$

All that remains is then 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 complex number.

Let \(\sqrt{3-4i\small\,\normalsize}=a+bi\small.\) So:

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

First we equate the imaginary part of this equation:

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

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

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

$$ \begin{eqnarray} a^2=-1\:\:\:\:&\small\textsf{or}\normalsize&\:\:a^2=4\\[6pt] \small\textsf{No solutions as }a\in\mathbb R\normalsize\:\:\:\:&\small\textsf{or}\normalsize&\:\:a=\pm 2\\[6pt] \end{eqnarray} $$

When \(a\!=\!\!-\!2\small,\) \(b\!=\!-\!\large\frac{2}{-2}\normalsize\!=\!1\) so one value of \(\sqrt{3-4i\small\,\normalsize\,}\) is \(-2+i\small.\)

When \(a\!=\!2\small,\) \(b\!=\!-\!\large\frac{2}{2}\normalsize\!=\!-\!1\) so the other value of \(\sqrt{3-4i\small\,\normalsize\,}\) is \(2-i\small.\)

If \(\raise 0.1pt{z}\) is a root, \(\raise 0.1pt{\overline{z}}\) is another root. Complex roots always exist in conjugate pairs.

So \(\raise 0.1pt{\overline{z}=1-2i}\) is another root.

Therefore \(\raise 0.1pt{z\!-\!(1\!+\!2i)}\) and \(\raise 0.1pt{z\!-\!(1\!-\!2i)}\) are factors of \(z^3-5z^2+11z-15\small.\)

Now we multiply these two linear factors to obtain a quadratic factor. This is made much easier if we use the regrouping method below:

$$ \begin{eqnarray} && \left(z\!-\!(1\!+\!2i)\right)\left(z\!-\!(1\!-\!2i)\right)\\[6pt] &=& (z-1-2i)(z-1+2i)\\[6pt] &=& ((z-1)-2i)((z-1)+2i)\\[6pt] &=& (z-1)^2 -4i^2 \\[6pt] &=& z^2-2z+1+4 \\[6pt] &=& z^2-2z+5 \\[6pt] \end{eqnarray} $$

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 \(\raise 0.1pt{\overline{z}=1+\sqrt{3}\tiny\,\normalsize i}\,\) is another root.

Therefore \(\raise 0.1pt{z\!-\!(1\!-\!\sqrt{3}\tiny\,\normalsize i)}\) and \(\raise 0.1pt{z\!-\!(1\!+\!\sqrt{3}\tiny\,\normalsize i)}\) are factors of \(z^4+3z^2+2z+12\small.\)

Now we multiply these two linear factors to obtain a quadratic factor:

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

Next we use polynomial long division, taking care to leave a space for the missing \(z^3\) term.

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

$$ \begin{eqnarray} z&=& \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\[6pt] &=& \frac{-2\pm\sqrt{2^{2}-4.1.3}}{2}\\[6pt] &=& \frac{-2\pm\sqrt{-8}}{2}\\[6pt] &=& \frac{-2\pm\sqrt{8}\tiny\,\normalsize i}{2}\\[6pt] &=& \frac{-2\pm2\sqrt{2}\tiny\,\normalsize i}{2}\\[6pt] &=& -1\pm\sqrt{2}\tiny\,\normalsize i \end{eqnarray} $$

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

(a)  Cartesian form: \(z=k\sqrt{3}-k\tiny\,\normalsize i\small.\) This is only 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 \(arg(z)\) and one for correctly combining them into polar form.

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

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

To find the principal argument, in the range \(-\pi\lt\theta\leq\pi\small,\) we use arctan:

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

So the polar form is:

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

Using our knowledge from the Functions topic that cosine is even and sine is odd, we might want to simplify this:

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

Note: Time permitting, it may be a wise idea in your exam to check that your polar form is correct by evaluating the cos and sin and checking that it simplifies back into the Cartesian form – but be careful to score out your check, in case the marker isn't sure that you knew where to stop!

(a)  To multiply two numbers in polar form we multiply their moduli and add their arguments, if necessary bringing the argument back into the range \(-\pi\lt\theta\leq\pi\) by adding or subtracting multiples of \(2\pi\) as required.

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

(b)  To divide two numbers in polar form we divide their moduli and subtract their arguments, if necessary bringing the argument back into the range \(-\pi\lt\theta\leq\pi\) by adding or subtracting multiples of \(2\pi\). That won't be necessary here.

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

First we express \(\raise 0.1pt{z}\) in polar form.

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

To find the principal argument, in the interval \(-\pi\lt\theta\leq\pi\small,\) we use arctan to find the related acute angle, and then adjust so that it is in the correct Argand diagram quadrant. Here, the number is in the third quadrant (bottom left), so we use \(\pi\) plus the acute angle and then subtract \(2\pi\) to bring it into the correct interval.

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

So \(z=\sqrt{2}\,\large\left(\normalsize cos\large\left(-\frac{3\pi}{4}\right)\normalsize\!+i\,sin\large\left(-\frac{3\pi}{4}\right)\right)\)

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

$$ [r(cos\,\theta\!\small+\normalsize\!i\,sin\,\theta)]^n = r^{n}(cos\,n\theta\!\small+\normalsize\!i\,sin\,n\theta) $$

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

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

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:

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

Let \(\raise 0.1pt{z^4=-\!1\!+\!i}\small.\)

First we express \(\raise 0.1pt{z^4}\) in polar form.

$$ \begin{eqnarray} \large\vert\normalsize z^4\large\vert\normalsize\ &=& \sqrt{(-1)^2+1^{2}}\\[6pt] &=& \sqrt{2} \end{eqnarray} $$

The argument is in the second quadrant on an Argand diagram, so

$$ \begin{eqnarray} arg(z^4) &=& \pi-tan^{-1}\LARGE\vert\small\frac{1}{-1}\LARGE\vert\normalsize \\[6pt] &=& \pi-\small\frac{\pi}{4}\normalsize \\[6pt] &=& \small\frac{3\pi}{4}\normalsize \\[6pt] \end{eqnarray} $$

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

So the fourth roots are as follows, not forgetting to allow for multiples of \(2\pi\small.\)

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

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

Now we apply de Moivre's Theorem:

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

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

To obtain separate expressions for each of these roots, we simply substitute each value of \(\raise 0.1pt{k}\) in turn.

When \(k=0\):

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

When \(k=1\):

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

When \(k=2\):

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

When \(k=3\):

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

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

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

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

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

To find the principal argument, in the range \(-\pi\lt\theta\leq\pi\small,\) we use arctan:

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

So the polar form is:

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

(b)  To find \(\raise 0.1pt{z^3}\) we use de Moivre's Theorem:

$$ [r(cos\,\theta\!\small+\normalsize\!i\,sin\,\theta)]^n = r^{n}(cos\,n\theta\!\small+\normalsize\!i\,sin\,n\theta) $$

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

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