Solution. (a) Multiply above and below by the conjugate of the denominator: \[\frac {2+3i}{1-i}=\frac {(2+3i)(1+i)}{(1-i)(1+i)} =\frac {2+2i+3i-3}{2}=-\frac {1}{2}+\frac {5}{2}i .\]

(b) Polar form is quicker than eight multiplications. Since \(1+i=\sqrt 2\,e^{i\pi /4}\), \[(1+i)^{8}=\left (\sqrt 2\right )^{8}e^{2\pi i}=16 .\]

(c) The powers of \(i\) repeat with period \(4\), and \(2026=4(506)+2\), so \(i^{2026}=i^{2}=-1\) and the expression is \(-1\).

Solution. Write \(8i=8e^{i\pi /2}\). The \(n\)-th roots of \(re^{i\theta }\) are \(r^{1/n}e^{i(\theta +2k\pi )/n}\) for \(k=0,\dots ,n-1\), so here \[z_k=2\exp \left [i\left (\frac {\pi }{6}+\frac {2k\pi }{3}\right )\right ], \qquad k=0,1,2 .\] In cartesian form \[z_0=\sqrt 3+i,\qquad z_1=-\sqrt 3+i,\qquad z_2=-2i .\] They lie on the circle of radius \(2\) at \(120^{\circ }\) apart, as the \(n\) roots of any number always do — the vertices of a regular \(n\)-gon.

Solution. (a) The condition says \(z\) is equidistant from \(1\) and \(-1\), so it is the perpendicular bisector of the segment joining them: the imaginary axis. Algebraically, with \(z=x+iy\), squaring gives \((x-1)^2+y^2=(x+1)^2+y^2\), hence \(x=0\).

(b) All points at distance \(3\) from \(2\): the circle of radius \(3\) centred at \(2+0i\).

(c) With \(z=x+iy\) the condition is \(x>y\), the open half plane below the line \(y=x\).

Solution. Using \(\left |w\right |^2=w\overline {w}\) and \(w+\overline {w}=2\operatorname {Re}(w)\), \begin {align*} \left |z_1+z_2\right |^2&=(z_1+z_2)\overline {(z_1+z_2)}\\ &=\left |z_1\right |^2+\left |z_2\right |^2+2\operatorname {Re}\left (z_1\overline {z_2}\right ). \end {align*}

Now \(\operatorname {Re}(w)\leq \left |w\right |\) for any \(w\), and \(\left |z_1\overline {z_2}\right |=\left |z_1\right |\left |z_2\right |\), so \[\left |z_1+z_2\right |^2\leq \left |z_1\right |^2+2\left |z_1\right |\left |z_2\right | +\left |z_2\right |^2=\left (\left |z_1\right |+\left |z_2\right |\right )^2 ,\] and taking non-negative square roots gives the inequality.

For the deduction, write \(z_1=(z_1-z_2)+z_2\) and apply it: \(\left |z_1\right |\leq \left |z_1-z_2\right |+\left |z_2\right |\), so \(\left |z_1\right |-\left |z_2\right |\leq \left |z_1-z_2\right |\). Exchanging \(z_1\) and \(z_2\) gives the same bound for \(\left |z_2\right |-\left |z_1\right |\), and the two together give the result.

Solution. De Moivre gives \((\cos \theta +i\sin \theta )^3=\cos 3\theta +i\sin 3\theta \). Expanding the left side with the binomial theorem and using \(i^2=-1\), \(i^3=-i\), \[\cos ^3\theta +3i\cos ^2\theta \sin \theta -3\cos \theta \sin ^2\theta -i\sin ^3\theta .\] Equating real and imaginary parts, \[\cos 3\theta =\cos ^3\theta -3\cos \theta \sin ^2\theta ,\qquad \sin 3\theta =3\cos ^2\theta \sin \theta -\sin ^3\theta .\] Substituting \(\sin ^2\theta =1-\cos ^2\theta \) turns the first into the familiar \(\cos 3\theta =4\cos ^3\theta -3\cos \theta \).