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19 Math II
z=(m^2-5m+6)+(m^2-3m)i

If 1 and z are real numbers and the coefficient before I is m 2-3m = 0, then m = 0 or m=3.

2, z is an imaginary number, then the coefficient m 2-3m before I is not equal to 0, so m is not equal to 0 or m is not equal to 3.

3, z is a pure imaginary number, then m 2-5m+6 = 0, and m 2-3m is not equal to 0.

M=2 or m=3, and m is not equal to 0 or 3.

So only m=2 can meet the requirements.

4.z=a+bi, where a stands for real part and b stands for imaginary part.

The coordinates (+,+) are in the first quadrant.

The coordinates (-,+) are in the second quadrant.

The coordinates (-,-) are in the third quadrant.

The coordinates (+,-) are in the fourth quadrant.

Z is in the second quadrant, so the real part m 2-5m+6: 0.

(m-2)(m-3)& lt; 0, and m (m-3) >; 0, so 2

Such m does not exist, so the absence of m makes the complex number z in the second quadrant.

I hope this is the right answer. I wish you progress in your study.