如果复数 $z_1=4+2 i, z_2=1-3 i$, 那么 $z_1 \cdot z_2=(\quad)$;

要计算复数 $ z_1 = 4 + 2i $ 和 $ z_2 = 1 - 3i $ 的乘积 $ z_1 \cdot z_2 $，我们使用复数乘法规则：

\[
z_1 \cdot z_2 = (4 + 2i)(1 - 3i)
\]

我们用分配律展开这个表达式：

\[
(4 + 2i)(1 - 3i) = 4 \cdot 1 + 4 \cdot (-3i) + 2i \cdot 1 + 2i \cdot (-3i)
\]

这将得到：

\[
= 4 - 12i + 2i - 6i^2
\]

由于 $i^2 = -1$，我们可以进一步简化：

\[
= 4 - 12i + 2i - 6(-1)
\]

\[
= 4 - 12i + 2i + 6
\]

\[
= 10 - 10i
\]

因此，复数的乘积是：

\[
z_1 \cdot z_2 = 10 - 10i
\]

Would you like more details or have any questions about this solution? Here are five more questions you might be interested in:

1. How do you divide complex numbers?
2. What is the geometric interpretation of complex number multiplication?
3. How do you find the magnitude of a complex number?
4. How do you add and subtract complex numbers?
5. How do you convert complex numbers between rectangular and polar form?

**Tip:** When multiplying complex numbers, always remember to apply the distributive property carefully and simplify using $ i^2 = -1 $.

Learn how to calculate the product of complex numbers z1 = 4 + 2i and z2 = 1 - 3i using the distributive property and simplification rules. This problem covers complex number multiplication, including the distributive property and simplification using i^2 = -1. Suitable for students in Grades 10-12.

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