
If the complex numbers iz, z and \[z + iz\]represents the three vertices of a triangle, then the area of the triangle is
A \[\dfrac{1}{2}\left| {z - 1} \right|\]
B \[{\left| z \right|^2}\]
C \[\dfrac{1}{2}{\left| z \right|^2}\]
D \[{\left| {z - 1} \right|^2}\]
Answer
533.1k+ views
Hint: Complex numbers are the numbers that are expressed in the form of \[a + ib\] where, a, b are real numbers and ‘i’ is an imaginary number. Here in the given question, we have the Vertices of triangle in complex form as; z, iz and \[z + iz\], hence here to find the area of the triangle we need to apply modulus to the terms and simplify to get the area.
Complete step by step answer:
Vertices of triangle in complex form: z, iz and \[z + iz\].
Let,
\[z = x + iy\]
\[iz = - y + ix\]
\[z + iz = x - y + i\left( {x + y} \right)\]
In cartesian form the vertices are:
\[\left( {x,y} \right);\left( { - y,x} \right);\left( {x - y,x + y} \right)\]
Hence, the area of triangle with respect to the vertices are:
\[Area = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
x&y&1 \\
{ - y}&x&1 \\
{x - y}&{x + y}&1
\end{array}} \right|\]
Evaluating the terms, as:
\[ = \dfrac{1}{2}\left[ {x\left( {x - x - y} \right) - y\left( { - y - x + y} \right) + 1} \right]\]
Now, combine the common terms and simplify it we get:
\[ = \dfrac{1}{2}\left[ { - xy + xy - {y^2} - {x^2}} \right]\]
As, area is positive, hence we have:
\[ = \dfrac{1}{2}\left[ {{x^2} + {y^2}} \right]\]
As, we have \[z = x + iy\] and \[{\left| z \right|^2} = {x^2} + {y^2}\], hence we get:
\[Area = \dfrac{1}{2}{\left| z \right|^2}\]
Therefore, option C is the right answer.
Additional information: Any number which is present in a number system such as positive, negative, zero, integer, rational, irrational, fractions, etc. are real numbers. The numbers which are not real are imaginary numbers. When we square an imaginary number, it gives a negative result are all imaginary numbers.
Note: We must know that the combination of both the real number and imaginary number is a complex number. And we must note the imaginary number is usually represented by ‘i’ or ‘j’, which is equal to \[\sqrt { - 1} \]. Therefore, the square of the imaginary number gives a negative value i.e., the values of \[{i^2} = - 1\] and \[i = \sqrt { - 1} \]. As we know, 0 is a real number and real numbers are part of complex numbers, therefore, 0 is also a complex number and is represented as \[0 + 0i\].
Complete step by step answer:
Vertices of triangle in complex form: z, iz and \[z + iz\].
Let,
\[z = x + iy\]
\[iz = - y + ix\]
\[z + iz = x - y + i\left( {x + y} \right)\]
In cartesian form the vertices are:
\[\left( {x,y} \right);\left( { - y,x} \right);\left( {x - y,x + y} \right)\]
Hence, the area of triangle with respect to the vertices are:
\[Area = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
x&y&1 \\
{ - y}&x&1 \\
{x - y}&{x + y}&1
\end{array}} \right|\]
Evaluating the terms, as:
\[ = \dfrac{1}{2}\left[ {x\left( {x - x - y} \right) - y\left( { - y - x + y} \right) + 1} \right]\]
Now, combine the common terms and simplify it we get:
\[ = \dfrac{1}{2}\left[ { - xy + xy - {y^2} - {x^2}} \right]\]
As, area is positive, hence we have:
\[ = \dfrac{1}{2}\left[ {{x^2} + {y^2}} \right]\]
As, we have \[z = x + iy\] and \[{\left| z \right|^2} = {x^2} + {y^2}\], hence we get:
\[Area = \dfrac{1}{2}{\left| z \right|^2}\]
Therefore, option C is the right answer.
Additional information: Any number which is present in a number system such as positive, negative, zero, integer, rational, irrational, fractions, etc. are real numbers. The numbers which are not real are imaginary numbers. When we square an imaginary number, it gives a negative result are all imaginary numbers.
Note: We must know that the combination of both the real number and imaginary number is a complex number. And we must note the imaginary number is usually represented by ‘i’ or ‘j’, which is equal to \[\sqrt { - 1} \]. Therefore, the square of the imaginary number gives a negative value i.e., the values of \[{i^2} = - 1\] and \[i = \sqrt { - 1} \]. As we know, 0 is a real number and real numbers are part of complex numbers, therefore, 0 is also a complex number and is represented as \[0 + 0i\].
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