Question

The complex number $z$ satisfying $|z + \overline z | + |z - \overline z | = 2$ and $|iz - 1| + |z - i| = 2$ is/are
A. $i$
B. $- i$
C. $\dfrac{1}{i}$
D. $\dfrac{1}{{{i^3}}}$

Answer 1

Hint : In the question they mentioned is/are, then two or more options are correct. Even all options may be correct. Since $z$ is a complex number then $z = x + iy$ . We need to find the value of $x$ and $y$ , by using the given conditions. Also we know that $|z| = \sqrt {{x^2} + {y^2}}$ . $|z|$ meaning $z$ lies in between -1 and 1.

Complete step-by-step answer :
Now given, $|z + \overline z | + |z - \overline z | = 2$ and $|iz - 1| + |z - i| = 2$ .
We need $\overline z$ and $iz$ and $z$ .
Now we know, $z = x + iy$
To find $iz = i(x + iy)$
$\Rightarrow = ix + {i^2}y$
We know that ${i^2} = - 1$
$\Rightarrow = ix + ( - 1)y$
$\Rightarrow = ix - y$
The conjugate of $z$ is $\overline z = x - iy$
Now take, $|z + \overline z | + |z - \overline z | = 2$
Substituting we get,
$\Rightarrow |x + iy + x - iy| + |x + iy - (x - iy)| = 2$
$\Rightarrow |x + iy + x - iy| + |x + iy - x + iy)| = 2$
Cancelling the terms we get,
$\Rightarrow |2x| + |2iy| = 2$
We know modulus of constant is constant only,
$\Rightarrow 2|x| + 2|i||y| = 2$
We know, $|i| = 1$
$\Rightarrow 2|x| + 2|y| = 2$
Divide by 2 on both side,
$\Rightarrow |x| + |y| = 1$ ------ (1)
Now take, $|iz - 1| + |z - i| = 2$
$\Rightarrow |ix - y - 1| + |x + iy - i| = 2$
$\Rightarrow |ix - (y + 1)| + |x + i(y - 1)| = 2$
We know $|z| = \sqrt {{x^2} + {y^2}}$ , applying for above equation we get,
$\Rightarrow \sqrt {{{(x)}^2} + {{[ - (y + 1)] }^2}} + \sqrt {{x^2} + {{(y - 1)}^2}} = 2$
$\Rightarrow \sqrt {{x^2} + {{(y + 1)}^2}} + \sqrt {{x^2} + {{(y - 1)}^2}} = 2$
Rearranging the terms,
$\Rightarrow \sqrt {{x^2} + {{(y + 1)}^2}} = 2 - \sqrt {{x^2} + {{(y - 1)}^2}}$
Squaring on the both sides,
$\Rightarrow {\left( {\sqrt {{x^2} + {{(y + 1)}^2}} } \right)^2} = {\left( {2 - \sqrt {{x^2} + {{(y - 1)}^2}} } \right)^2}$
We know ${(a - b)^2} = {a^2} + {b^2} - 2ab$ , applying on right hand side of the above equation,
$\Rightarrow {x^2} + {(y + 1)^2} = {2^2} + {\left( {\sqrt {{x^2} + {{(y - 1)}^2}} } \right)^2} - 2 \times 2\sqrt {{x^2} + {{(y - 1)}^2}}$
We know ${(a + b)^2} = {a^2} + {b^2} + 2ab$ using this we get,
$\Rightarrow {x^2} + {y^2} + 1 + 2y = {2^2} + {x^2} + {(y - 1)^2} - 4\sqrt {{x^2} + {{(y - 1)}^2}}$
$\Rightarrow {x^2} + {y^2} + 1 + 2y = 4 + {x^2} + {y^2} + 1 - 2y - 4\sqrt {{x^2} + {{(y - 1)}^2}}$
Cancelling terms ${x^2}$ and ${y^2}$ we get,
$\Rightarrow 1 + 2y = 4 + 1 - 2y - 4\sqrt {{x^2} + {{(y - 1)}^2}}$
Cancelling 1 on both sides,
$\Rightarrow 2y + 2y = 4 - 4\sqrt {{x^2} + {{(y - 1)}^2}}$
$\Rightarrow 4y - 4 = - 4\sqrt {{x^2} + {{(y - 1)}^2}}$
$\Rightarrow 4(y - 1) = - 4\sqrt {{x^2} + {{(y - 1)}^2}}$
$\Rightarrow (y - 1) = \sqrt {{x^2} + {{(y - 1)}^2}}$
Squaring on both sides,
$\Rightarrow {(y - 1)^2} = {\left( {\sqrt {{x^2} + {{(y - 1)}^2}} } \right)^2}$
$\Rightarrow {(y - 1)^2} = {x^2} + {(y - 1)^2}$
Cancelling ${(y - 1)^2}$ term on both side,
$\Rightarrow 0 = {x^2}$ Or ${x^2} = 0$
Hence, $x = 0$
To find $y$ , from equation (1) we have $\Rightarrow |x| + |y| = 1$
Substituting we get,
$\Rightarrow |0| + |y| = 1$
$\Rightarrow |y| = 1$
Modulus of any variable lies between -1 and 1 ,
$\Rightarrow y = \pm 1$
Now substituting in $z = x + iy$
That is $x = 0,y = 1$ we get,
$\Rightarrow z = 0 + i(1) = i$
Now substituting $x = 0,y = - 1$ , we get,
$\Rightarrow z = 0 + i( - 1) = - i$
We get both $i$ and $- i$ so both options (a) and (b) are correct.
Let’s check the other options
Now take option (c), that is $\dfrac{1}{i}$ . Rationalizing we get:
$\Rightarrow \dfrac{1}{i} \times \dfrac{i}{i} = \dfrac{i}{{{i^2}}} = \dfrac{i}{{ - 1}} = - i$ ( ${i^2} = - 1$ )
The simplified form of $\dfrac{1}{i}$ is $- i$ .
Hence option (c) is also correct.
Now take option (d) that is $\dfrac{1}{{{i^3}}}$ .
$\dfrac{1}{{{i^3}}} = \dfrac{1}{{ - i}}$ ( ${i^3} = {i^2}.i = - i$ )
Rationalizing we get,
$\dfrac{1}{{ - i}} \times \dfrac{i}{i} = \dfrac{i}{{ - ({i^2})}} = \dfrac{i}{{ - ( - 1)}} = i$ ( ${i^2} = - 1$ )
Simplified form of $\dfrac{1}{{{i^3}}}$ is $i$ .
Hence option (d) is also correct.
So, we can see that from above all the options are correct.
So, the correct answer is “Option A,B,C and D”.

Note : This problem seems to be long and has a lot of calculation. All we did is basic math and if we have some basic knowledge about complex numbers we can solve this problem easily. Remember these things in complex number $z = x + iy$ , ${i^2} = - 1$ and $|z| = \sqrt {{x^2} + {y^2}}$ . These are the basic known formulas in complex numbers