Question

If z and w are two complex numbers such that $\left| z \right|\le 1$, $\left| w \right|\le 1$ and$\left| z-iw \right|=\left| z-i\bar{w} \right|=2$. Then z is equal to
(1) $1$ or $i$
(2) $i$ or $-i$
(3) $1$ or $-1$
(4) $i$ or $-1$

Answer 1

First we will write what is given in the question and according to that we will solve our question further. All the properties of complex numbers should be clear to you and all the formulas. A complex number contains one real part and one imaginary part. The value of the real part or the imaginary part can also be equal to zero.

Complete step-by-step solution:
When two real and imaginary numbers are added to each other then they form a complex number. A number is said to be a complex number if it can be written in the form of $a+ib$. The value of i is $\sqrt{-1}$.
In the above question, it is given that
z and w are two complex numbers.
$\left| z \right|\le 1$and $\left| w \right|\le 1$
It is also given that, the value of $\left| z-iw \right|$ and the value of $\left| z-i\bar{w} \right|$ is equal to 2
That is,$\left| z-iw \right|=2$
$\left| z-i\bar{w} \right|=2$
Now we will solve our question.
From above
$\left| z-iw \right|=2$………eq(1)
On squaring eq(1) on both sides, we get
${{\left| z-iw \right|}^{2}}={{(2)}^{2}}$
After solving this, we get
$(z-iw)(\bar{z}+i\bar{w})=4$
( from the identity ${{\left| z \right|}^{2}}=z\bar{z}$)
Now we will multiply them and the following result will be obtained
$z\bar{z}+izw-i\bar{w}\bar{z}+(-{{i}^{2}}w\bar{w})=4$
(And we know that $-{{i}^{2}}=1$)
So the above equation becomes
$z\bar{z}+izw-i\bar{w}\bar{z}+(w\bar{w})=4$……..eq(2)
Also in the above question, it is given that
$\left| z-i\bar{w} \right|=2$……..eq(3)
On squaring eq(3) on both the side, we get
${{\left| z-i\bar{w} \right|}^{2}}={{(2)}^{2}}$
On solving this equation, the following result is obtained
$(z-i\bar{w})(\bar{z}+iw)=4$
( from the identity ${{\left| z \right|}^{2}}=z\bar{z}$)
So the above equation becomes
$z\bar{z}+ziw-i\bar{w}\bar{z}+(-{{i}^{2}}w\bar{w})=4$
(And we know that $-{{i}^{2}}=1$)
So the above equation becomes
$z\bar{z}+ziw-i\bar{w}\bar{z}+(w\bar{w})=4$……….eq(4)
On subtracting eq(2) from eq(4), we get
$z\bar{z}+ziw-i\bar{w}\bar{z}+(w\bar{w})-z\bar{z}-izw+iw\bar{z}-w\bar{w}=0$
On solving this equation, we get
$(z+\bar{z})(w-\bar{w})=0$
The first case will be when $w-\bar{w}=0$, and w is a purely real number
The second case will be when $z-\bar{z}=0$, and z is a purely imaginary number
In the above question, it is given that
$\left| z-iw \right|\le 2$
So, $\left| z \right|+\left| iw \right|\le 2$
Also in the question, it is given that $\left| z \right|\le 1$and also $\left| w \right|\le 1$
But in the question, it is given that $\left| z-iw \right|=2$
This is possible only when
$\left| z \right|=1$
$\left| w \right|=1$
Z is a purely imaginary number and also the value of $\left| z \right|=1$
value of z will be $i$ or $-i$
So the correct answer is (2).

Note: If we have to add two complex numbers with each other, then the easiest way to add them is to first add the real part and then add the imaginary part. Conjugation of a complex number means if we change the sign between the real and imaginary part. If we have to divide two complex numbers then their conjugate is used.