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`        If |z-2-i|=4 then find the minimum value of |3z-6+15i| plz reply fast as possible`
one year ago

```							|z-2-i|=4z-2-i=+4,-4z=6+i,(-2+i)1.z=6+i|3(6+i)-6+15i|=|12+18i|2.z= -2+I|3(-2+i)-6+15i|=|12-18i|=MINIMUM
```
one year ago
```							|z – 2 – i| = 4  or  |z – (2 + i)| = 4  Above equation represents a circle centred at (2,1) with radius 4 units. Eqn of circle is (x – 2)2 + (y – 2)2 = 16            ................(1)Minimum value of  |3z-6+15i| means minimum value of 3|z – (2 – 5i)|. First we’ll find the minimum value of |z – (2 – 5i)| & multiply it by 3.It denotes least distance from a point on the circle & point (2, – 5).(2, – 5) lies below the circle in coordinate axes system.Line through center (2,1) & (2, – 5) is parallel to y-axis &its eqn is x = 2          …................(2)points of intersection of (1) & (2) are (2,5) & (2, – 3).By geometry, (2, –3) is nearest to (2, –5) of all points on the circle.  Distance between (2, –3) & (2, –5) is 2 units, i.e., minimum value of |z – (2 – 5i)| is 2 and hence the minimum value of |3z-6+15i| = 3|z – (2 – 5i)| = 3*2                                                      = 6
```
one year ago
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