Question

The area of the triangle whose vertices are represented by the complex numbers\[0\], \[z\], \[z{e^{i\alpha }}\],\[\left( {0 < \alpha < \pi } \right)\]equals [AMU\[2002\]]
A) \[\dfrac{1}{2}|z{|^2}cos\alpha \]
B) \[\dfrac{1}{2}|z{|^2}\sin \alpha \]
C) \[\dfrac{1}{2}|z{|^2}\sin \alpha \cos \alpha \]
D) \[\dfrac{1}{2}|z{|^2}\]

Answer 1

Hint: in this question we have to find the modulus of one vertices of triangle. In order to find this we have to use formula of area of triangle and expand the determinant to get area of triangle. First write each vertex as combination of real and imaginary number.



Formula Used:Area of triangle is given by
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
Where
A is area of triangle
\[{a_i}\] is a real part of vertices
\[{b_i}\] is a imaginary part of vertices



Complete step by step solution:Given: Three vertices of triangle
Vertex of triangle is given as \[0\], \[z\], \[z{e^{i\alpha }}\]
We know that
\[z = x + iy\]
Where x is real part of z and iy is imaginary part of z here i(iota) makes iy imaginary
\[z{e^{i\alpha }} = (x + iy)(\cos \alpha + i\sin \alpha )\]
\[z{e^{i\alpha }} = (x\cos \alpha - y\sin \alpha ) + i(y\cos \alpha + x\sin \alpha )\]
Now area of triangle is given by
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
Where
A is area of triangle
\[{a_i}\] is a real part of vertices
\[{b_i}\] is a imaginary part of vertices
Expand determinant to get area of triangle
\[A = \dfrac{1}{2}({a_1}({b_2} - {b_3}) - {a_2}({b_1} - {b_3}) + {a_3}({b_1} - {b_2}))\]
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}0&0&1\\x&y&1\\{(xcos\alpha - ysin\alpha )}&{(ycos\alpha + xsin\alpha )}&1\end{array}} \right|\]
Expand the determinant to get area of triangle
\[A = \dfrac{1}{2}[xycos\alpha + {x^2}sin\alpha - xycos\alpha + {y^2}sin\alpha ]\;\;\]
\[A = \dfrac{1}{2}sin\alpha ({x^2} + {y^2})\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]

\[{\left| z \right|^2} = {x^2} + {y^2}\]
\[A = \dfrac{1}{2}|z{|^2}sin\alpha \]



Option ‘B’ is correct



Note: We have to remember that complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part.
Imaginary part is known as iota. Square of iota is equal to negative one.