怎么解这道题---**Example/Problem:** **Example 7.** Dividing f(z) by z - i, where i = sqrt(-1), we obtain the remainder i and dividing it by z + i, we get the remainder 1 + i. Find the remainder upon the division of f(z) by z^2 + 1.

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We have a polynomial division problem. We're given that when f(z) is divided by z minus i, the remainder is i. When f(z) is divided by z plus i, the remainder is 1 plus i. We need to find the remainder when f(z) is divided by z squared plus 1. Notice that z squared plus 1 factors as z minus i times z plus i. Now we apply the remainder theorem. This theorem states that when polynomial f of z is divided by z minus a, the remainder equals f of a. From our given conditions, we can determine that f of i equals i, and f of negative i equals 1 plus i. Since z squared plus 1 has degree 2, the remainder must be a linear polynomial of the form a z plus b. Now we set up the system of equations. We express the division as f of z equals z squared plus 1 times q of z plus R of z, where R of z equals a z plus b. Since this equation holds for all z, it must hold for z equals i and z equals negative i. Substituting z equals i gives us a i plus b equals i. Substituting z equals negative i gives us negative a i plus b equals 1 plus i. This gives us our system of linear equations. Now we solve the system of equations. We have a i plus b equals i and negative a i plus b equals 1 plus i. Adding these equations eliminates the a i terms: 2b equals 1 plus 2i, so b equals 1 plus 2i over 2. Subtracting the second equation from the first gives us 2a i equals negative 1, so a equals i over 2. Let's verify: substituting back into the first equation gives i over 2 times i plus 1 plus 2i over 2 equals i, which checks out. The second equation also verifies correctly. Here's our final answer. The remainder polynomial is R of z equals i over 2 times z plus 1 plus 2i over 2, which can be simplified to iz plus 1 plus 2i all over 2. Let's verify this is correct. When z equals i, we get i times i plus 1 plus 2i over 2, which equals negative 1 plus 1 plus 2i over 2, giving us i. When z equals negative i, we get 1 plus 1 plus 2i over 2, which equals 1 plus i. Both check out perfectly. This method works for any similar polynomial division problem by applying the remainder theorem and solving the resulting system of equations.