If Im$$\left( \dfrac{z-i}{2i} \right) =0$$, then the locus of z is:
A. x-axis
B. y-axis
C. The line x=y
D. The line x+y+1=0
Answer
664.2k+ views
Hint: In this question it is given that the imaginary part of $$\left( \dfrac{z-i}{2i} \right)$$ is zero, then we have to find the locus of z. So to find the locus we have to first rationalise the given fraction i.e, we have to omit ‘i’ from the denominator and after that locus of Complex Numbers is obtained by letting z = x + iy and simplifying the expressions.
Complete step-by-step solution:
Let us consider z=x+iy, where i is the square root of -1.
Therefore here it is given that,
$$\text{Im} \left( \dfrac{z-i}{2i} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{z-i}{2i} \times \dfrac{i}{i} \right) =0$$ [multiplying numerator and denominator by i]
$$\Rightarrow \text{Im} \left( \dfrac{(z-i)\times i}{2i\times i} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{iz-i^{2}}{2i^{2}} \right) =0$$
Since as we know that i is the square root of -1 therefore,
So by putting the value in the above equation we get,
$$\text{Im} \left( \dfrac{iz-\left( -1\right) }{2\times \left( -1\right) } \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{iz+1}{-2} \right) =0$$
Now since as we know that z=x+iy, therefore we get,
$$ \text{Im} \left( \dfrac{i\left( x+iy\right) +1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{ix+i^{2}y+1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{ix+\left( -1\right) y+1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{ix-y+1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{-ix+y-1}{2} \right) =0$$ [taking ‘-‘ in the numerator]
$$\Rightarrow \text{Im} \left\{ \dfrac{y-1}{2} +i\left( \dfrac{-x}{2} \right) \right\} =0$$
As we know that if any term multiplied with ‘i’ then this term is called the imaginary part, and since in the question it is given that imaginary part is zero, so we can write,
$$ \dfrac{-x}{2} =0$$
$$\Rightarrow -x=0$$
$$\Rightarrow x=0$$
Therefore x=0 is the equation of the locus, which is also an equation of y-axis.
Hence the correct option is option B.
Note: While solving complex number related problems you need to know that a complex number is a number that can be expressed in the form a + ib, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation $$i^{2}=-1$$. For the complex number a +ib, a is called the real part, and b is called the imaginary part.
Complete step-by-step solution:
Let us consider z=x+iy, where i is the square root of -1.
Therefore here it is given that,
$$\text{Im} \left( \dfrac{z-i}{2i} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{z-i}{2i} \times \dfrac{i}{i} \right) =0$$ [multiplying numerator and denominator by i]
$$\Rightarrow \text{Im} \left( \dfrac{(z-i)\times i}{2i\times i} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{iz-i^{2}}{2i^{2}} \right) =0$$
Since as we know that i is the square root of -1 therefore,
So by putting the value in the above equation we get,
$$\text{Im} \left( \dfrac{iz-\left( -1\right) }{2\times \left( -1\right) } \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{iz+1}{-2} \right) =0$$
Now since as we know that z=x+iy, therefore we get,
$$ \text{Im} \left( \dfrac{i\left( x+iy\right) +1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{ix+i^{2}y+1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{ix+\left( -1\right) y+1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{ix-y+1}{-2} \right) =0$$
$$\Rightarrow \text{Im} \left( \dfrac{-ix+y-1}{2} \right) =0$$ [taking ‘-‘ in the numerator]
$$\Rightarrow \text{Im} \left\{ \dfrac{y-1}{2} +i\left( \dfrac{-x}{2} \right) \right\} =0$$
As we know that if any term multiplied with ‘i’ then this term is called the imaginary part, and since in the question it is given that imaginary part is zero, so we can write,
$$ \dfrac{-x}{2} =0$$
$$\Rightarrow -x=0$$
$$\Rightarrow x=0$$
Therefore x=0 is the equation of the locus, which is also an equation of y-axis.
Hence the correct option is option B.
Note: While solving complex number related problems you need to know that a complex number is a number that can be expressed in the form a + ib, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation $$i^{2}=-1$$. For the complex number a +ib, a is called the real part, and b is called the imaginary part.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

