Solution. (a). \((2-3i)-(1+2i)=2-3i-1-2i=1-5i\).

(b). \((3-i)^{2}=9-6i+i^{2}=9-6i-1=8-6i\).

(c). Square the numerator first: \((1+2i)^{2}=1+4i+4i^{2}=-3+4i\). Then multiply above and below by the conjugate \(2-i\): \[\frac {(-3+4i)(2-i)}{(2+i)(2-i)}=\frac {-6+3i+8i-4i^{2}}{4+1} =\frac {-2+11i}{5}=-\frac {2}{5}+\frac {11}{5}i .\]

Solution. (a). \(z_1z_2=(8+2i)(2+i)=16+8i+4i-2=14+12i\), so \[\frac {14+12i}{3+i}=\frac {(14+12i)(3-i)}{9+1} =\frac {42-14i+36i+12}{10}=\frac {54+22i}{10}=\frac {27}{5}+\frac {11}{5}i .\]

(b). \(z_1-z_2=6+i\) and \(z_1+z_3=11+3i\), so \[\frac {6+i}{11+3i}=\frac {(6+i)(11-3i)}{121+9} =\frac {66-18i+11i+3}{130}=\frac {69-7i}{130}=\frac {69}{130}-\frac {7}{130}i .\]

(c). \(z_1-2z_3=8+2i-6-2i=2\) and \(3z_2=6+3i\), so \[\frac {2^{2}}{(6+3i)^{2}}=\frac {4}{36+36i-9}=\frac {4}{27+36i} =\frac {4(27-36i)}{27^{2}+36^{2}}=\frac {108-144i}{2025} =\frac {4}{75}-\frac {16}{225}i .\] Note how much work part (c) saves by simplifying \(z_1-2z_3\) to the real number \(2\) before squaring anything.

Solution. Put \(z=p+qi\), so that \(z^{2}=4+3i\). Then \[(p+qi)^{2}=p^{2}-q^{2}+2pqi=4+3i .\] Comparing real and imaginary parts gives two equations, \[p^{2}-q^{2}=4 \qquad \text {and}\qquad 2pq=3 .\] From the second, \(q=\frac {3}{2p}\); substituting into the first, \[p^{2}-\frac {9}{4p^{2}}=4 \implies 4p^{4}-16p^{2}-9=0 .\] This is a quadratic in \(p^{2}\): \(\left (2p^{2}-9\right )\left (2p^{2}+1\right )=0\). Since \(p\) is real, \(p^{2}\) cannot be negative, so \(2p^{2}=9\) and \(p=\pm \frac {3}{\sqrt {2}}=\pm \frac {3\sqrt {2}}{2}\). Then \(q=\frac {3}{2p}\) takes the same sign as \(p\), giving \(q=\pm \frac {\sqrt {2}}{2}\).

Hence \[z=\pm \left (\frac {3\sqrt {2}}{2}+\frac {\sqrt {2}}{2}i\right ).\] Check: \(\left (\frac {3\sqrt {2}}{2}+\frac {\sqrt {2}}{2}i\right )^{2} =\frac {18}{4}-\frac {2}{4}+2\cdot \frac {3\sqrt {2}}{2}\cdot \frac {\sqrt {2}}{2}i=4+3i\).