The (complex) solutions of the equation \( 2 z^{2}+2 i z-5=0 \) are \( \ldots \) \( \begin{array}{l}z_{1}=-1.5-0.5 i \text { and } z_{2}=1.5+0.5 i \\ z_{1}=-0.5-1.5 i \text { and } z_{2}=0.5+1.5 i \\ z_{1}=-1.5-0.5 i \text { and } z_{2}=1.5-0.5 i \\ z_{1}=1.5+0.5 i \text { and } z_{2}=1.5-0.5 i\end{array} \)

We start with the quadratic equation   2z² + 2iz – 5 = 0. Using the quadratic formula   z = [–b ± √(b² – 4ac)] / (2a), with a = 2, b = 2i, and c = –5, we get   z = [–2i ± √((2i)² – 4·2·(–5))] / (2·2). Step 1. Compute b²:   (2i)² = 4i² = –4. Step 2. Compute the discriminant:   b² – 4ac = –4 – [4·2·(–5)] = –4 + 40 = 36. Step 3. Take the square root of the discriminant:   √36 = 6. Thus, the solutions are   z = (–2i ± 6) / 4. This gives two solutions:   z₁ = (–2i + 6) / 4 = (6 – 2i)/4 = (3 – i)/2,   z₂ = (–2i – 6) / 4 = (–6 – 2i)/4 = –(6 + 2i)/4 = –(3 + i)/2. Now we convert these into decimal form:   (3 – i)/2 = 1.5 – 0.5i,   –(3 + i)/2 = –1.5 – 0.5i. Looking at the answer choices provided, the option   z₁ = –1.5 – 0.5i and z₂ = 1.5 – 0.5i matches our solutions (note that the order of the solutions doesn’t matter). Thus, the correct answer is:   z₁ = –1.5 – 0.5i and z₂ = 1.5 – 0.5i.