The greatest common factor of 4 and 8 is () and the least common multiple is

The greatest common factor of 4 and 8 is () and the least common multiple is

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C programming language: input two positive integers m and N, find their greatest common divisor and least common multiple
Input two positive integers m and N to find their greatest common divisor and least common multiple. Note: the greatest common divisor is also called the greatest common factor, which refers to the greatest one of several integers' common factors; the common multiple of two integers is called their common multiple, and the smallest positive integer is called their least common multiple; integer numbers: ")、printf("\nthe greatest common divisor is … 、printf("\nthe least common multiple is … The running effect of the program should be similar to that shown in Figure 1. 35 & nbsp; 15 in Figure 1 is the content input from the keyboard
#include int main() { int m, n; int m_ cup, n_ Cup, & nbsp; res; & nbsp; / * divisor, & nbsp; divisor, & nbsp; remainder * / & nbsp; printf (& quot; enter & nbsp; two & nbsp; integer): & n & quot;); & nb
|If Z | = 2Z + 9i, what is the complex z =?
Z = a + bi, a, B are real numbers
|Z | is a real number
So 2A + 2bi + 9i is also a real number
So 2B + 9 = 0
b=-9/2
therefore
|z|=2a
So a ^ 2 + B ^ 2 = 4A ^ 2
a^2=b^2/3=27/4
a=±3√3/2
Because | Z | > = 0, so a > = 0
So z = 3 √ 3 / 2-9i / 2
Given the complex number Z1, Z2, the module satisfying Z1 is root 7 + 1, the module of Z2 is root 7-1, the module of z1-z2 is 4, find Z1 / Z2
I can't use a + bi, C + Di in the end
There is another way to answer that is Z1, Z2 are the two sides of a rectangle
So why can Z1 / Z2 = (radical 7 + 1 / radical 7-1) * I do this?
Because the modules of Z1, Z2, (z1-z2) just constitute Pythagorean theorem, Z1 and Z2 can be regarded as two sides of a rectangle
Using the coordinate method, if Z2 is the x-axis and Z2 is the y-axis, then Z1 is (0, root 7 + 1) and Z2 is (root 7-1,0);
So Z1, Z2 can be expressed as Z1 = (root 7 + 1) * I, Z2 = root 7-1
Therefore, Z1 / Z2 = (radical 7 + 1 / radical 7-1) * I
If the complex Z is equal to 1 plus the root 2I, then what is the square of Z minus 2Z
If it's equal to - 3, you can count it as normal
However, in most problems, the complex number is rarely squared, usually multiplied by its own conjugate module. It looks like a square,
Can we talk about the process
Given that the complex number Z = a ^ 2 + A-2 + (a ^ 2-3a + 2) I is a pure imaginary number, then the value of the real number a?
a^2 + a - 2 = 0 => a = 1 or a = -2
A = 1, a ^ 2 - 3A + 2 = 0
a = -2
If the complex Z satisfies 1 / z = Z / (3z-10), | Z|=
1/Z=Z/(3Z-10)
That is: Z & sup2; = 3z-10
z²-3z+10=0
∴ z=（3±i*√31）/2
|Z|= √10
Given that the imaginary number Z satisfies Z & sup2; + 49 / Z & sup2; ∈ R, the graph of the set of corresponding points of Z on the complex plane is obtained
Let z = a + bi, substitute, divide, make the imaginary part of the molecule equal to zero, then we can get the relationship between a and B. (note that a and B cannot be zero at the same time.)
If the complex Z satisfies the equation Z ^ 2 + 2Z + 4 = 0, then z =?
Z = (- 2 + / - root 16-4) / 2 (root formula)
Z = - 1 + radical 3I / - 1 - radical 3I
What does 1 / (1 + I) + I / 2 equal?
1/(1+i)+i/2=（1-i）/2+i/2=1/2