1.求下列复数z的实部与虚部、共轭复数、模与辐角:-|||-(1) 1/(3+2i);-|||-(2) dfrac (1)(i)-dfrac (3i)(1-i);-|||-(3) dfrac ((3+4i)(2-5i))(2i);-|||-(4) https:/img.zuoyebang.cc/zyb_f4c780ed2ac2d743fc2449876c657113.jpg^8-(4i)^21+i.

步骤 1：计算 $\dfrac {1}{3+2i}$
首先，将复数分母实部化，即乘以共轭复数：
$$\dfrac {1}{3+2i} = \dfrac {1}{3+2i} \cdot \dfrac {3-2i}{3-2i} = \dfrac {3-2i}{9+4} = \dfrac {3}{13} - \dfrac {2}{13}i$$
步骤 2：计算 $\dfrac {1}{i}-\dfrac {3i}{1-i}$
首先，将每个分式分母实部化：
$$\dfrac {1}{i} = \dfrac {1}{i} \cdot \dfrac {-i}{-i} = -i$$
$$\dfrac {3i}{1-i} = \dfrac {3i}{1-i} \cdot \dfrac {1+i}{1+i} = \dfrac {3i(1+i)}{2} = \dfrac {3i+3i^2}{2} = \dfrac {-3+3i}{2}$$
然后，将两个结果相减：
$$-i - \dfrac {-3+3i}{2} = \dfrac {3}{2} - \dfrac {5}{2}i$$
步骤 3：计算 $\dfrac {(3+4i)(2-5i)}{2i}$
首先，计算分子：
$$(3+4i)(2-5i) = 6 - 15i + 8i - 20i^2 = 6 - 7i + 20 = 26 - 7i$$
然后，将结果除以 $2i$：
$$\dfrac {26 - 7i}{2i} = \dfrac {26 - 7i}{2i} \cdot \dfrac {-i}{-i} = \dfrac {-26i - 7i^2}{-2} = \dfrac {-26i + 7}{-2} = -\dfrac {7}{2} + 13i$$
步骤 4：计算 ${1}^{8}-4{i}^{21}+i$
首先，计算 $i^{21}$：
$$i^{21} = (i^4)^5 \cdot i = 1^5 \cdot i = i$$
然后，将结果代入原式：
$$1^8 - 4i + i = 1 - 3i$$