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Question

If z1 and z2 are complex numbers such that Rez1=z1-1, Rez2=z2-1, and argz1-z2=π6, then Imz1+z2 is equal to:


A

23

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B

23

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C

13

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D

32

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Solution

The correct option is A

23


Explanation for the correct option.

Step 1. Assume two complex numbers.

Let z1=x1+iy1 and z2=x2+iy2 be two complex numbers.

The real part of z1 is Rez1=x1 and the real part of z2 is Rez2=x2.

Step 2. Form the equation using the given information Rez1=z1-1.

It is given that Rez1=z1-1, so substitute Rez1=x1 and z1=x1+iy1 .

Rez1=z1-1x1=x1+iy1-1x1=x1-1+iy1x1=x1-12+y12

Now, square both sides.

x12=x1-12+y122x12=x12-2x1+1+y12-y12=-2x1+1y12=2x1-1...(1)

Step 3. Form the equation using the given information Rez2=z2-1.

It is given that Rez2=z2-1, so substitute Rez2=x2 and z2=x2+iy2 .

Rez2=z2-1x2=x2+iy2-1x2=x2-1+iy2x2=x2-12+y22

Now, square both sides.

x22=x2-12+y222x22=x22-2x2+1+y22-y22=-2x2+1y22=2x2-1...(2)

Step 4. Form the equation using the given information argz1-z2=π6.

In the given equation argz1-z2=π6 substitute z1=x1+iy1 and z2=x2+iy2.

argx1+iy1-x2-iy2=π6argx1-x2+i(y1-y2)=π6tan-1y1-y2x1-x2=π6argx+iy=tan-1yxy1-y2x1-x2=tanπ6y1-y2x1-x2=13...(3)

Step 5. Find the value of Imz1+z2.

Subtract equation 2 from equation 1.

y12-y22=2x1-1-2x2+1y1-y2y1+y2=2x1-x2a2-b2=(a-b)(a+b)y1+y2=2x1-x2y1-y2y1+y2=21y1-y2x1-x2y1+y2=2113Usingequation3y1+y2=23

Now, Imz1+z2 is given as

Imz1+z2=Imx1+iy1+x2+iy2=Imx1+x2+iy1+y2=y1+y2Imx+iy=y=23y1+y2=23

So, the value of Imz1+z2 is 23.

Hence, the correct option is A.


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