Question
Question: In complex numbers \[z\mathop z\limits^\\_ = 0\] if and only if (A) \(\operatorname{Re} \left( z ...
In complex numbers z\mathop z\limits^\\_ = 0 if and only if
(A) Re(z)=0
(B) z=0
(C) Im(z)=0
(D) Re(z)=Im(z)
Solution
To solve this problem we have to assume z=a+ib and after that we put this in z\mathop z\limits^\\_ = 0 and solve them.In z=a+ib, Real part of z is a and imaginary part of z is b And ∣z∣=a2+b2.Using these definitions and formulas we try to get the answer.
Complete step-by-step answer:
We assume that z=a+ib
And modulus of z
∣z∣=a2+b2
So now given z\mathop z\limits^\\_ = 0
As we know that z\mathop z\limits^\\_ = {\left| z \right|^2}
Now so from above given equation we can write
z\mathop z\limits^\\_ = {\left| z \right|^2} = 0
∵∣z∣=a2+b2
So ∣z∣2=a2+b2
And we know that
∣z∣2=0
So ∣z∣2=a2+b2=0
This condition is possible only when a=0 and b=0
So from this we can say z=0+i0
So from this we say that
Re(z)=0 as well as Im(z)=0
And z=0 because the real part and imaginary part are both zero.
Now as we see Re(z)=0 as well as Im(z)=0
From this we can say Re(z)=Im(z)=0
So all four options are the correct answer.
So, the correct answer is “All options”.
Note: A complex number z=x+iy is a purely real if its imaginary part is 0, i.e. Im(z) = 0 and purely imaginary if its real part is 0 i.e. Re(z) = 0.Two complex numbers z1=x1+iy1 and z2=x2+iy2 are equal, if x1=x2 and y1=y2 i.e. Re(z1) = Re(z2) and Im(z1) = Im(z2).
Order relation “greater than’’ and “less than” are not defined for complex numbers.
Important identities:
1. Additive identity z + 0 = z = 0 + z
Here, 0 is an additive identity.
2. Multiplicative identity: z×1 = z = 1×z
3. Conjugate of Complex Number: Let z=x+iy, if ‘i’ is replaced by (−i), then said to be conjugate of the complex number z and it is denoted by \mathop z\limits^\\_ , i.e. \mathop z\limits^\\_ = x - iy.