When taking the value of the real number M, the complex number z 2 5m 3 6 m 1 i is 2 imaginary numbe

z=(2 +5m-3)-(6 -m-1)i When 6m -m-1=(3m+1)(2m-1)=0, i.e., m=-1 3 or m=1 2, it is a real number;

When 6m -m-1 = (3m + 1) (2m -1) ≠ 0, i.e. m≠-1 3 and m≠1 2, it is an imaginary number;

When 2 +5m-3 = (m+3)(2m-1) = 0 and 6m -m-1 = (3m+1)(2m-1) ≠ 0, i.e., m=-3, it is a pure imaginary number.

This question mainly examines the definitions of imaginary numbers and pure imaginary numbers, so it can be calculated according to the definitions of imaginary numbers and pure imaginary numbers. The specific methods are as follows: (1) Definition of imaginary numbers:

If a bi is an imaginary number and a,b r, then b ≠ 0. So if z is imaginary, then 6m -m-1≠0 solves m≠1 2 and m≠-1 3That is, when m≠1 2 and m≠-1 3, z is an imaginary number.

2) Definition of pure imaginary number: if a+bi is a pure imaginary number and a,b r, then b≠0,a=0So if z is a pure imaginary number, then there is 2m +5m-3=0 and 6m -m-1≠0, and the solution is m=-3.

That is, when m=-3, z is a pure imaginary number. Hope it helps!

If the imaginary number is a pure imaginary number, the real part is 0, i.e.

2m +5m-3=0 then the solution is m=1 2 or 3 and z is an imaginary number, the imaginary part is not 0, 6m -m-1≠0 solution is m≠1 2 or 1 3 when the above complex numbers are pure imaginary numbers

m≠-3 when it is an imaginary number

1) When m-6=0, undefeated i.e., m=6, the complex number z is a real number;

2) When m-6≠0, i.e., m≠6, the complex number z is an imaginary number;

3) When m+1=0, and m-6≠0, and m=-1, the plural lead Chazhou z is a pure imaginary number

1. Argument of real numbers.

That is, the imaginary part is 0, ie.

m squared - 3m = 0

m(m-3)=0m=0 or.

m=3 imaginary number: the imaginary part is not 0

Namely. m(m-3)≠0

m≠0 and. m≠3

Pure imaginary numbers. That is, the real part is 0

The imaginary part is not 0m squared - 5m + 6 = 0

m-2)(m-3)=0

m = 2 or m = 3

When m=3, the imaginary part is also 0, and Z is 0, which is still a real number with the stove.

Therefore. m=2

Solution: Z=(m -2m+m i)-[4+(5m-6)i]=(m -2m-4)+(m -5m+6)i

Real number: imaginary coefficient = 0

m²-5m+6=0

m-2)(m-3)=0

m = 2 or m = 3

Imaginary number: imaginary coefficient≠0 m≠2 and m≠3

Pure imaginary number: imaginary coefficient≠0 real part = 0

m²-2m-4=0

m-1)²=5

m=1±√5

Zero: imaginary part = 0, real part = 0, known from the above problem solving process, no solution.

z=（m2-5m+6）+（m2-3m）i

z=(m-2)(m-3) +m(m-3)i is an imaginary number, then the imaginary part cannot be 0, so when m is not equal to 0 and m is not equal to 3, the original formula is an imaginary number pure imaginary number means: the real part is zero, so when m 2, it is a pure imaginary number ---

z = (m2-5m +6) + (m2-3m) I when m2-3m = m(m-3) = 0, such that m = 0 or 3:00 and z = 6 or 0, z is a real number;

square meters = meters (m-3) ≠0, meters ≠0 and m≠3, in z is an imaginary number; p> when 6 = (m-2) (m-3) = 0 in square meters of 5 m, and 3 m in square meters = m (m-3) ≠ 0, m = 2, that is, and z = -2i, z is a pure imaginary number.

z=[(m-3)(m+2) (m+3)]+m+3)(m-5)iFor fractions, m≠-3

For real numbers, the imaginary part is zero. So: m=5

imaginary number, then the imaginary part is not zero. So: m≠-3 and m≠5 is a pure imaginary number, then the real part is zero. So: m=3 or m=-2

Answer. Imaginary numbers:

6m²－m－1≠0

3m+1)(2m－1)≠0

i.e. m≠ 1 3 and m≠1 2

is a pure imaginary number: 6m m 1≠0

and 2m + 5m 3 = 0

Solve the second one.

2m－1)(m+3)=0

m=1 2 or 3

Combined with the condition that z is an imaginary number, m= 3 satisfies the topic.

Two conditions are satisfied, one is 2m 2+5m-3=0, that is, m=1 2 and m=-3, and the other is 6m 2-m-1≠0, that is, m≠1 2 and m≠-1 3, when the two conditions are combined, m=-3, z is a pure imaginary number.