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Question

If z=x+iy, then area of the triangle whose vertices are points z,iz,z+iz is


A

12z2

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B

14z2

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C

z2

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D

32z2

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Solution

The correct option is A

12z2


Explanation for the correct option.

Step 1. Find the vertices of the triangle.

It is given that z=x+iy is a complex number and its modulus is |z|=x2+y2

Now iz is defined as:

iz=i(x+iy)=ix+i2y=ix+-1yi2=-1=-y+ix

And the complex number z+iz is defined as:

z+iz=x+iy+-y+ix=x-y+ix+y

As z, iz, and z+iz are the vertices of a triangle, so the coordinates of the vertices are: x,y,-y,x,x-y,x+y.

Step 2. Find the area of the triangle.

The area of the triangle having vertices as x,y,-y,x,x-y,x+y is given as:

Area=12xy1-yx1x-yx+y1=12xx-x+y-y-y-x-y+1-yx+y-xx-y=12xx-x-y-y(-y-x+y)+-xy-y2-x2+xy=12-xy+xy-y2-x2=12-x2+y2=12x2+y2=12x2+y22=12z2|z|=x2+y2

So, the area of the triangle whose vertices are points z,iz,z+iz is 12z2.

Hence, the correct option is A.


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