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Question: If Im\[\left( \dfrac{z-i}{2i} \right) =0\], then the locus of z is: A. x-axis B. y-axis C. The...

If Im(z−i2i)=0\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

Explanation

Solution

Hint: In this question it is given that the imaginary part of (z−i2i)\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,
Im(z−i2i)=0\text{Im} \left( \dfrac{z-i}{2i} \right) =0
⇒Im(z−i2i×ii)=0\Rightarrow \text{Im} \left( \dfrac{z-i}{2i} \times \dfrac{i}{i} \right) =0 [multiplying numerator and denominator by i]
⇒Im((z−i)×i2i×i)=0\Rightarrow \text{Im} \left( \dfrac{(z-i)\times i}{2i\times i} \right) =0
⇒Im(iz−i22i2)=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,
Im(iz−(−1)2×(−1))=0\text{Im} \left( \dfrac{iz-\left( -1\right) }{2\times \left( -1\right) } \right) =0
⇒Im(iz+1−2)=0\Rightarrow \text{Im} \left( \dfrac{iz+1}{-2} \right) =0
Now since as we know that z=x+iy, therefore we get,
Im(i(x+iy)+1−2)=0\text{Im} \left( \dfrac{i\left( x+iy\right) +1}{-2} \right) =0
⇒Im(ix+i2y+1−2)=0\Rightarrow \text{Im} \left( \dfrac{ix+i^{2}y+1}{-2} \right) =0
⇒Im(ix+(−1)y+1−2)=0\Rightarrow \text{Im} \left( \dfrac{ix+\left( -1\right) y+1}{-2} \right) =0
⇒Im(ix−y+1−2)=0\Rightarrow \text{Im} \left( \dfrac{ix-y+1}{-2} \right) =0
⇒Im(−ix+y−12)=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,
−x2=0\dfrac{-x}{2} =0
⇒−x=0\Rightarrow -x=0
⇒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 i2=−1i^{2}=-1. For the complex number a +ib, a is called the real part, and b is called the imaginary part.