如何用C显示i的负数平方根 [英] How to Print Out the Square Root of A Negative Number with i Displayed in C

问题描述

我试图弄清楚如果正好是负数（由用户输入）时如何显示数字的平方根，如果是，请正确显示显示的 i。当我执行普通的 sqrt 函数时，结果总是类似于-1。＃IND。当我尝试使用双复数变量时，正数或负数都将变得干净。

I am trying to figure out how to display the square root of a number if it happens to be negative (as it is entered by the user), and if so, display it correctly with the "i" displayed as well. When I do the normal sqrt function, the result is always something like -1.#IND. When I tried using the double complex variables, the positive numbers nor the negative numbers would come out clean.

下面是我的代码；评论是我的目标。用户输入4个变量，可以是正整数或负整数。

Below is my code; the comments are what my goal is. The 4 num variables are entered by the user and can be any integer, positive or negative.

 //  Display the square root of each number.  Remember that the user can enter negative numbers and 
//  will need to find the negative root with the "i" displayed.
printf("\nThe square root of %d is %.4f", num1, sqrt(num1));
printf("\nThe square root of %d is %.4f", num2, sqrt(num2));
printf("\nThe square root of %d is %.4f", num3, sqrt(num3));
printf("\nThe square root of %d is %.4f", num4, sqrt(num4));

推荐答案

如果使用浮点运算，则可以使用内置的复杂实用程序，例如：

If you're working with floating point you can use the built-in complex utilities, e.g.:

#include <complex.h>
#include <stdio.h>

int main(void)
{
    double complex num = -4.0;
    double complex s = csqrt(num);

    printf("%.2f + %.2fi\n", creal(s), cimag(s));
}

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