C language topic, input two positive integers m and N, find its greatest common divisor and least common multiple It's OK to give a prompt. There's no need to write the code. Please note that the prompt is not too simple

C language topic, input two positive integers m and N, find its greatest common divisor and least common multiple It's OK to give a prompt. There's no need to write the code. Please note that the prompt is not too simple

Greatest common divisor: (round Division)
Adjust the order so that M > = n
loop
M = M% n; if M = 0, then n is the greatest common divisor, jumping out of the loop
N = n% m; if n = 0, then M is the greatest common divisor, jumping out of the loop
next
After finding out the greatest common divisor g, we use M * n / g to get the least common multiple
C language from the keyboard input two positive integers m and N, the greatest common divisor and the least common multiple
#include
void main()
{
int m,n,r,x;
scanf("%d,%d",&m,&n);
x=m*n;
while(n!=0)
{
r=m%n;
m=n;
n=r;
}
printf("%d %d\n",m,x/m);
}
Given that the complex Z satisfies the equation Z ^ 2-3 | Z | + 2 = 0, then z =?
Let z = x + iy be substituted into the equation:
x^2+2ixy-y^2-3√(x^2+y^2)+2=0
If the imaginary part is 2XY = 0, then x = 0 or y = 0
Real part = x ^ 2-y ^ 2-3 √ (x ^ 2 + y ^ 2) + 2 = 0
When x = 0, the real part = - y ^ 2-3 | y | + 2 = 0, then: | y | ^ 2 + 3 | y | - 2 = 0, | y | = (- 3 + √ 17) / 2, then: y = (- 3 + √ 17) / 2, (3 - √ 17) / 2
When y = 0, the real part = x ^ 2-3 | x | + 2 = 0, then: | x | = 1 or 2, then: x = 1, - 1,2, - 2,
So there are six solutions
Let Z1 = 1-3i, Z2 = 3-2i, then which quadrant is the corresponding point of Z1 / Z2 in the complex plane
Z1 / Z2 = (9-7i) / 13 so quadrant 4
It's all in the fourth quadrant
Z1/Z2=(1-3i)(3+2i)/13=9/13-7i/13
In the fourth quadrant
On the equation Z2 - (a + I) Z - (I + 2) = 0 (a ∈ R) of complex number Z, (1) if the equation has a real number solution, find the value of a; (2) prove with the method of counter proof that for any real number a, the original equation cannot have pure imaginary root
(1) If the equation has a real solution, let z = m ∈ R, then M2 - (a + I) m - (I + 2) = 0, that is, m2-am-2 + (- m-1) I = 0, m2-am-2 = 0, and - M-1 = 0, M = - 1, a = 1. (2) suppose that the original equation has a pure imaginary root, let z = Ni, n ≠ 0, then there is (Ni) 2 - (a + I) Ni - (a + 2) I = 0, and the entire equation can be - N2 + n-2 + (- an-1) I = 0, n − N2 & nbsp; + n & nbsp; − 2 & nbsp; = & nbsp; 0 & nbsp; &For ①, the discriminant △ 0, the equation ① has no solution, so the equation system has no solution, so the hypothesis is not tenable, so the original equation can not have pure imaginary roots
How to define the complex number in mathematics?
In the range of real numbers, negative numbers have no square root, but in some scientific calculations, it is necessary to use the square root of negative numbers, so I is used to express sqrt (- 1), that is, the square root of - 1. Complex numbers are numbers containing I, including real numbers and other complex numbers (that is, numbers containing I). - 1 + 10i1010i are complex numbers
A number can be expressed as a + bi
Where a and B are real numbers and the square of I is - 1
How many solutions of the equation Z ^ 2-3 | Z | + 2 = 0 in the complex set?
Let z = x + Yi get x ^ 2-y ^ 2 + 2xyi-3 √ (x ^ 2 + y ^ 2) + 2 = 0 [x ^ 2-y ^ 2-3 √ (x ^ 2 + y ^ 2) + 2] + 2xyi = 0x ^ 2-y ^ 2-3 √ (x ^ 2 + y ^ 2) + 2 = 0, ① 2XY = 0, ② get x = 0 or y = 0x = 0, - y ^ 2-3 | y | + 2 = 0, y ^ 2 + 3 | y | - 2 = 0, there are two solutions
We all know that the square of Hai is - 1, so how much is the cubic fourth power and the fifth power of I pushed down all the time? For example, how much is the 63rd power of I? Is there a trick
The first power of I is I, the second power is - 1, the third power is - I, and the fourth power is 1. If I push back, the fourth power of 4N + 1 is I, the fourth power of 4N + 2 is - 1, the fourth power of 4N + 3 is - I, and the fourth power of 4N + 3 is 1 (n can only be positive integers). There are only four answers. For example, the 63rd power of I, because 63 = 4 * 15 + 3, is the same result as the third power of I, which is - I
Solving complex equation | Z-2 | - z = 1 + 3I
|z-2|=z+1+3i
Because the left is a real number, so the right is also a real number, so z = a-3i
|z-2|=a+1
|a-2-3i|=a+1
Square: (A-2) ^ 2 + 9 = (a + 1) ^ 2
Expansion: - 4A + 4 + 9 = 2A + 1
6a=12
A=2
That is Z = 2-3i
How to calculate complex number in Matlab mathematical calculation
Detailed input steps are required
The specific codes and results in MATLAB are shown in the figure below
Real part + imaginary part I or real part + imaginary part J