We have a complex number z with absolute value 1, and need to find the minimum of |iz + z̄| + |2iz - z̄|. Since |z| = 1, we can represent z as e^(iθ) on the unit circle. This parametric form will help us analyze the expression systematically.
Now we substitute z = e^(iθ) into our expression. We have z equals cosine theta plus i sine theta, and z conjugate equals cosine theta minus i sine theta. For the first term iz + z̄, we get i times z plus z conjugate, which simplifies to cosine theta minus sine theta times the factor 1 plus i. For the second term 2iz - z̄, we get a more complex expression with both real and imaginary parts.
这是一个关于复数模长的最值问题。题目要求在单位圆约束下，求两个复数模长之和的最小值。我们将使用三角函数的参数化方法来解决这个问题。
由于复数z满足模长为1的约束，我们可以将其参数化为单位圆上的点。设z等于余弦theta加i正弦theta，那么z的共轭就是余弦theta减i正弦theta。计算iz和相关表达式，我们得到两个复数的具体形式。
现在我们计算两个复数的模长。对于第一个表达式，我们使用模长的定义，展开并利用三角恒等式，最终得到根号下2减正弦2theta。对于第二个表达式，我们计算模长的平方，展开后简化得到5加4倍正弦2theta的平方根。
我们将问题转化为关于参数t的单变量函数优化问题。通过求导并令导数为零，我们找到临界点。解方程得到t等于27/20。这个值超出了t的定义域，所以我们需要在端点处寻找最小值。
由于临界点超出了定义域，我们需要检查边界点。在t等于负1时，函数值为根号3加1；在t等于1时，函数值为4。比较可知，最小值在t等于负1处取得，对应的最小值是根号3加1。
We transform the problem into a single-variable optimization. Setting t equals sine of 2 theta, we get a function g of t. Taking the derivative and setting it to zero, we find the critical point at t equals 27 over 20. However, this value exceeds the domain boundary of negative 1 to 1, so we must check the boundary values instead.
Now we evaluate the function at the boundary points. At t equals negative 1, we get square root of 3 plus 1, which is approximately 2.732. At t equals 1, we get 4. Since square root of 3 plus 1 is less than 4, the minimum value is 3, achieved when sine of 2 theta equals negative 1.