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Question

If z-4z=2, then the maximum value of z is


A

3+1

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B

5+1

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C

2

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D

2+2

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Solution

The correct option is B

5+1


Explanation for the correct option.

Step 1. Form a compound inequality.

For any two complex number a and b a-ba-b.

So for complex number z and 4z

z-4zz-4zz-4z2z-4z=2

Now, let z=r>0, then r-4r2 and so the compound inequality becomes

-2r-4r2

Step 2. Solve the left inequality.

From the compound inequality -2r-4r2, the left inequality is r-4r-2, the inequality can be written as:

r2-4-2rr2+2r-40

The roots of the quadratic equation r2+2r-4=0 is given as:

r=-2±22-4×1×-42×1=-2±4+162=-2±252=-1±5

So, the solution of the inequality r2+2r-40 is r-1-5 and r-1+5.

But as r>0, so r-1-5 is rejected.

Thus the solution from the ldeft inequality is r-1+5...1

Step 3. Solve the right inequality.

From the compound inequality -2r-4r2, the right inequality is r-4r2, the inequality can be written as:

r2-42rr2-2r-40

The roots of the quadratic equation r2-2r-4=0 is given as:

r=-(-2)±(-2)2-4×1×-42×1=2±4+162=2±252=1±5

So, the solution of the inequality r2-2r-40 is 1-5r1+5 .

But as r>0, so the solution from the right inequality is 0<r1+5...(2).

Step 4. Find the combined solution and find the maximum value of z.

The solution from the left inequality is r-1+5 and from the right inequality is 0<r1+5. So the combined solution is:

-1+5r1+5.

As r=z, so -1+5z1+5.

Thus the maximum value of z is 5+1.

Hence, the correct option is B.


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