It is known that Z 1 = 1-3i, Z 2 = 6-8i. If 1z + 1z 1 = 1z 2, find the value of Z

It is known that Z 1 = 1-3i, Z 2 = 6-8i. If 1z + 1z 1 = 1z 2, find the value of Z

∵z1=1-3i，z2=6-8i，1z+1z1=1z2，∴z=z1•z2z1−z2=(1−3i)(6−8i)1−3i−(6−8i)=−18−26i−5+5i=(−18−26i)(−5−5i)(−5+5i)(−5−5i)=−40+220i25+25=-45+225i．

Given that Z1 = a + 3I (a is greater than zero) and the module of Z = 5, find Z2

Is this Z2 a conjugate complex of Z1?
Explanation
The membrane of Z1 is 5
So a ^ 2 + 3 ^ 2 = 5 ^ 2
a^2=16
A = 4 or a = - 4
Z2 = 4-3i or Z2 = - 4-3i

The reciprocal of Z1 = 1 + 3I, Z2 = 3 + I, Z1, Z, Z2 is an arithmetic sequence, and Z is obtained

The reciprocal of Z1, Z, Z2 is an arithmetic sequence
2/z
=1/z1+1/z3
=1/(1+3i)+1/(3+i)
=(1-3i)/(1+9)+(3-i)/(9+1)
=(2-2i)/5
z=10/(2-2i)
=5/(1+i)
=5(1-i)/(1+1)
So z = (5-5i) / 2

If Z1 = 1 + 2I, Z2 = 1-3i, z = Z1 * Z2, then Z is not =?

Z = Z1 * Z2 = (1 + 2I) (1-3i) = 1-I + 6 = 7-i, Z not = I-7

fifteen

15

It is known that one side of an isosceles triangle is 24, and the waist is twice the bottom. Find the perimeter of the triangle

When the base is 24, waist length is 24 * 2 = 48, circumference is 24 + 48 * 2 = 120
When the waist length is 24, the base length is 24 / 2 = 12, and the circumference is 24 * 2 + 12 = 60

Given that the radius of circle O is 1, P is a point outside circle O, PA is tangent to circle O at point a, PA = 1, AB is the chord of circle O, and ab = 2, radical 2, then the length of Pb is
There are two answers, one and five

In the Title Ab should be equal to the root 2, otherwise no do
There are two answers
Because point B can be on the same side or on the opposite side of point P
When B and P are on the same side, the quadrilateral OAPB is a square, so Pb = 1
When B and P are opposite, the quadrilateral OAPB is a parallelogram, so Pb = root 5

As shown in the figure, Mn is the edge ab of △ ABC, and the two fixed points on AC find a point P on BC to make the perimeter of △ PMN shortest
Hope to see the picture above

What about the picture

As shown in the figure, in ladder ABCD, ad ∥ BC and BD are diagonals, and the median EF intersects BD at point O. if fo-eo = 3, BC-AD equals ()
A. 4B. 6C. 8D. 10

∵ EF is trapezoid, ABCD is median line, ∥ EF ∥ BC ∥ ad. ∥ ob = OD. ∥ BC = 2of, ad = 2oe. ∥ BC-AD = 2 (fo-eo) = 2 × 3 = 6

It is proved that if f (x) = ax + B, then f [(X &; + X &;) / 2] = [f (X &;) + F (X &;)] / 2

Left end = a * [(x1 + x2) / 2] + B = (ax1 + AX2 + 2b) / 2
Right end = [(ax1 + b) + (AX2 + b)] / 2 = (ax1 + AX2 + 2b) / 2
Left end = right end, so the original formula holds