Solveeit Logo

Question

Question: If \[\dfrac{{{{\left| {(a + ib)} \right|}^2}}}{{(a - ib)}} - \dfrac{{{{\left| {(a - ib)} \right|}^2}...

If (a+ib)2(aib)(aib)2(a+ib)=x+iy\dfrac{{{{\left| {(a + ib)} \right|}^2}}}{{(a - ib)}} - \dfrac{{{{\left| {(a - ib)} \right|}^2}}}{{(a + ib)}} = x + iy then find the value of xx.

Explanation

Solution

The given question is based on complex numbers. It can be solved easily by applying the following property for complex numbers:
z.z=z2z.\overline z = {\left| z \right|^2}
We have to expand the equation using the given property and then simplify to arrive at the value of xx.

Complete step by step solution:
Complex numbers are the numbers that are expressed in the form of a+iba + ib where, aa,bb are real numbers and ‘ii’ is an imaginary number called “iota”. For example, 4+6i4 + 6i is a complex number, where 44 is a real number (Re) and 6i6i is an imaginary number (Im).
An imaginary number is usually represented by ‘ii’ or ‘jj’, which is equal to 1\sqrt { - 1} . Therefore, the square of the imaginary number gives a negative value.
Let z=a+ibz = a + ib be a complex number.
The Modulus of z is represented by z\left| z \right|.
Mathematically, z=a2+b2\left| z \right| = \sqrt {{a^2} + {b^2}}
The conjugate of “zz” is denoted by z\overline z .
Mathematically, z=aib\overline z = a - ib
We can solve the given equation as follows:
(a+ib)2(aib)(aib)2(a+ib)=x+iy\dfrac{{{{\left| {(a + ib)} \right|}^2}}}{{(a - ib)}} - \dfrac{{{{\left| {(a - ib)} \right|}^2}}}{{(a + ib)}} = x + iy
Using the property z.z=z2z.\overline z = {\left| z \right|^2}, we get,
(a+ib)(a+ib)(aib)(aib)(aib)(a+ib)=x+iy\dfrac{{(a + ib)\overline {(a + ib)} }}{{(a - ib)}} - \dfrac{{(a - ib)\overline {(a - ib)} }}{{(a + ib)}} = x + iy
Converting the sign as per conjugate rule z=aib\overline z = a - ib, we get,
(a+ib)(aib)(aib)(aib)(a+ib)(a+ib)=x+iy\dfrac{{(a + ib)(a - ib)}}{{(a - ib)}} - \dfrac{{(a - ib)(a + ib)}}{{(a + ib)}} = x + iy
Simplifying the equation by dividing, we get,
(a+ib)(aib)=x+iy(a + ib) - (a - ib) = x + iy
Opening the brackets, we get,
a+iba+ib=x+iya + ib - a + ib = x + iy
2ib=x+iy2ib = x + iy
Writing the equation in form of complex number z=a+ibz = a + ib, we get,
x+iy=0+i(2b)x + iy = 0 + i(2b)
Hence, from the above equation, we can conclude that the value of xx corresponds to 00.

Therefore, x=0x = 0.

Note:
The complex number is a mixture of a real number and an imaginary number. The main application of these numbers is to represent periodic motions such as water waves, alternating current, light waves, etc., which rely on sine or cosine waves, etc.