In the complex plane, rotate the vector corresponding to the complex number 2-radical 5I by 2 / 2 in the counter clockwise direction. What is the modulus of the complex number corresponding to the obtained vector

In the complex plane, rotate the vector corresponding to the complex number 2-radical 5I by 2 / 2 in the counter clockwise direction. What is the modulus of the complex number corresponding to the obtained vector

When a vector is rotated, its module is invariant,
m'=m0=sqrt(4+5)=3
Module equals three

In the complex plane, the vector corresponding to the complex number 3 + I is oz. if the vector oz rotates 60 ° counterclockwise around the origin of the coordinate, the complex number corresponding to the vector oz 'is oz______ ．

The complex number corresponding to the vector is (3 + I) (cos60 ° + isin60 °) = (3 + I) (12 + 32i) = 2I, so the answer is 2I

In the complex plane, the vector corresponding to 3-radical 3 I rotates π / 3 clockwise, and the corresponding complex number is
A 2 root 3
B - 2 radical 3I
C radical 3-3i
D 3 + radical 3I
Does the vector rotate around the origin? How can I work out that it coincides with the y-axis?

The answer is B, which coincides with the negative half axis of Y. the meaning of the question is that the vector rotates around the origin. The 3-radical 3 I corresponds to the vector OA = (3, - radical 3). The angle between the 3-radical 3 I and the X axis is 30 degrees. Therefore, after rotating 60 degrees clockwise, it coincides with the negative half axis of Y axis. Obviously, the real part is 0 and the imaginary part is negative

A sufficient and necessary condition for making z-a / Z + a (a > 0) a pure imaginary number

Since a > 0, then a is a real number
z²－a²/z²＋a²=[(m²－b²－a²)＋2mbi]/[(m²－b²＋a²)＋2mbi]
The real part of this complex denominator is (M & sup2; - B & sup2; - A & sup2;) (M & sup2; - B & sup2; + A & sup2;) + 4m & sup2; B & sup2; = (M & sup2; - B & sup2;) & sup2; - (A & sup2;) & sup2; + 4m & sup2; B & sup2; = (M & sup2; + B & sup2;) & sup2; - (A & sup2;) & sup2; = 0, that is, M & sup2; + B & sup2; = A & sup2;, the modulus of complex Z is equal to a
The reduced imaginary part of the complex number is (M & sup2; - B & sup2; - A & sup2;) (- 2MB) + (M & sup2; - B & sup2; + A & sup2;) (2MB) = (2MB) (2a & sup2;) = 4mba & sup2; ≠ 0, that is, m ≠ 0 and B ≠ 0
So far, the necessary and sufficient condition is that the module of Z is a, and the real part and imaginary part of complex Z are not equal to 0