00:01
In this question, we need to match the elements of column 1 with the elements of column 2.
00:06
The part a is saying that minimum value of of modulus z minus 2 plus iota plus modulus z minus 2 iota plus 1 will be.
00:36
We know that modulus z1 minus z2 is less than equals to modulus z1 plus modulus z2.
00:54
We can say that let's say modulus z1 is equals to modulus z minus 2 plus iota and modulus z 2 is equals to modular z minus 2 iota plus 1 substitute these values in this equation let's say this was equation 1 so we get modulus z minus 2 plus iota plus modulus z minus 2 aorta plus 1 greater than equals to modulus minus 2 plus iota minus z minus 2 iota plus 1 when we further solve it we get minus 2 iota plus 2 iota minus 1 and this will be equals to minus 3 plus 3 aota and this will be equal to under root of minus 3 square plus 3 square is equals to 3 root 2.
02:39
Therefore the correct match for the first part is a and s.
03:01
Now let's come to the part b.
03:14
The part b is saying that minimum value of of modulus z1 minus z2 where modulus z1 plus 1 minus iota is equal to 13 and modulus z3 minus 3 minus 4 aota is equal to 2 we can observe that these two are the equations of circle therefore modulus z 1 plus 1 minus iota is equal to 13 is a circle and its center is minus 1 comma 1 and its radius is 3.
04:27
Similarly, modulus z3 minus 3 minus 4 aota is equals to 2 is the circle.
04:37
Its center is 3 .4 and radius is 2.
04:56
Based on this information we can draw these circles and the circle is shown below.
05:11
So these are the two circles.
05:15
Let's say this point is c1.
05:18
This is c2, this point is a and this point is b.
05:30
The coordinates of c1 are minus 1 .1 and the coordinates of c2 are 3 .4.
05:45
Therefore we can write minimum value of modulus z1 minus z2 is equal to ab and this will be equals to c1b minus c1c2 plus c2a.
06:22
From the figure we can see that c1b is equal to 13 and c2a is equal to 2...