Question

If $|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}$ , then $\dfrac{{{z_1}}}{{{z_2}}}$ is
(1) purely real
(2) purely imaginary
(3) zero of purely imaginary
(4) neither real nor imaginary

Answer 1

Hint: A complex number is a number that can be expressed in the form $a + ib$ , where $a$ and $b$ are real numbers, and $i$ is a symbol called the imaginary unit, and satisfying the equation ${i^2} = - 1$ . First we analyze the square part and omit the same term and then we get the required answer.

Complete step-by-step answer:
First we collect the given data $|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}$
We use the property $|{z_1} + {z_2}{|^2} = ({z_1} + {z_2})(\overline {{z_1}} + \overline {{z_2}} )$ and we get
$({z_1} + {z_2})(\overline {{z_1}} + \overline {{z_2}} ) = |{z_1}{|^2} + |{z_2}{|^2}$
Now we know that $({z_1} + {z_2})(\overline {{z_1}} + \overline {{z_2}} ) = |{z_1}{|^2} + |{z_2}{|^2} + {z_1}\overline {{z_2}} + {z_2}\overline {{z_1}} $
Use this in the above equation and we get
$ \Rightarrow |{z_1}{|^2} + |{z_2}{|^2} + {z_1}\overline {{z_2}} + {z_2}\overline {{z_1}} = |{z_1}{|^2} + |{z_2}{|^2}$
Now we omit the similar terms and we get
$ \Rightarrow {z_1}\overline {{z_2}} + {z_2}\overline {{z_1}} = 0$
Now we divide both sides of the above equation by ${z_2}\overline {{z_2}} $ , we get
$ \Rightarrow \dfrac{{{z_1}\overline {{z_2}} + {z_2}\overline {{z_1}} }}{{{z_2}\overline {{z_2}} }} = \dfrac{0}{{{z_2}\overline {{z_2}} }}$
Now we simplifying above equation and omit the similar terms and we get
$ \Rightarrow \dfrac{{{z_1}\overline {{z_2}} }}{{{z_2}\overline {{z_2}} }} + \dfrac{{{z_2}\overline {{z_1}} }}{{{z_2}\overline {{z_2}} }} = 0$
$ \Rightarrow \dfrac{{{z_1}}}{{{z_2}}} + \dfrac{{\overline {{z_1}} }}{{\overline {{z_2}} }} = 0$
To finding the answer let us consider $\dfrac{{{z_1}}}{{{z_2}}} = x + iy$ and $\dfrac{{\overline {{z_1}} }}{{\overline {{z_2}} }} = x - iy$
Use this in above equation and we get
$ \Rightarrow (x + iy) + (x - iy) = 0$
$ \Rightarrow 2x = 0$
From the above equation we say that the real part of $\left( {\dfrac{{{z_1}}}{{{z_2}}}} \right)$ is zero.
i.e., $\operatorname{Re} \left( {\dfrac{{{z_1}}}{{{z_2}}}} \right) = 0$
From the above condition we can say that the $\dfrac{{{z_1}}}{{{z_2}}}$ is purely imaginary.
So, the correct answer is “Option 2”.

Note: We try to remember the property of complex numbers. We remember that the double bar of a complex number is always the complex number itself. The part $\operatorname{Re} (z)$ is equal to $z + \overline z $ , where $z = a + ib$ and $\overline z = a - ib$ . If we get the real part of a function is zero then always remember this function is purely imaginary. We use the property of complex number $|{z_1} + {z_2}{|^2} = ({z_1} + {z_2})(\overline {{z_1}} + \overline {{z_2}} )$ to simplify the method of solving and easy to understand.