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Question: If \(z\) is a complex number such that \({\mathop{\rm Re}\nolimits} (z) = {\mathop{\rm Im}\nolimits}...

If zz is a complex number such that Re(z)=Im(z){\mathop{\rm Re}\nolimits} (z) = {\mathop{\rm Im}\nolimits} (z), then
(A) Re(z2)=0{\mathop{\rm Re}\nolimits} ({z^2}) = 0 (B) Im(z2)=0{\mathop{\rm Im}\nolimits} ({z^2}) = 0 (C) Re(z2)=Im(z2){\mathop{\rm Re}\nolimits} ({z^2}) = {\mathop{\rm Im}\nolimits} ({z^2}) (D) Re(z2)=−Im(z2){\mathop{\rm Re}\nolimits} ({z^2}) = - {\mathop{\rm Im}\nolimits} ({z^2})

Explanation

Solution

Hint- Assume z=x+iyz = x + iy as a complex number, where xx and yy are real numbers. To solve this question we need to find the real part and imaginary part of z2{z^2}.

Complete Step by step solution:
Let z=x+iyz = x + iy, where x≠0x \ne 0 and y≠0y \ne 0
Also,
xx is real part of zz i.e. Re(z){\mathop{\rm Re}\nolimits} (z) and,
yy is imaginary part i.e. Im(z){\mathop{\rm Im}\nolimits} (z)
As given in the question,
Re(z)=Im(z){\mathop{\rm Re}\nolimits} (z) = {\mathop{\rm Im}\nolimits} (z)
⇒x=y\Rightarrow x = y equation (1)
Now, we take square of zz,
z2=(x+iy)2{z^2} = {(x + iy)^2}
z2=x2−y2+2ixy{z^2} = {x^2} - {y^2} + 2ixy
Here again we have real part and imaginary part as Re(z2){\mathop{\rm Re}\nolimits} ({z^2}) and Im(z2){\mathop{\rm Im}\nolimits} ({z^2}).
For Re(z2){\mathop{\rm Re}\nolimits} ({z^2}),
Re(z2)=x2−y2{\mathop{\rm Re}\nolimits} ({z^2}) = {x^2} - {y^2}
⇒Re(z2)=x2−x2\Rightarrow {\mathop{\rm Re}\nolimits} ({z^2}) = {x^2} - {x^2} [from equation (1)]
⇒Re(z2)=0\Rightarrow {\mathop{\rm Re}\nolimits} ({z^2}) = 0 equation (2)
And forIm(z2){\mathop{\rm Im}\nolimits} ({z^2}),
Im(z2)=2xy{\mathop{\rm Im}\nolimits} ({z^2}) = 2xy
Which can’t be 00 as x≠0x \ne 0 and y≠0y \ne 0
So Im(z2)≠0{\mathop{\rm Im}\nolimits} ({z^2}) \ne 0 equation (3)
From equation (2) and equation (3) we have some conclusions as
Re(z2)=0{\mathop{\rm Re}\nolimits} ({z^2}) = 0, Im(z2)≠0{\mathop{\rm Im}\nolimits} ({z^2}) \ne 0 and therefore Re(z2)≠Im(z2){\mathop{\rm Re}\nolimits} ({z^2}) \ne {\mathop{\rm Im}\nolimits} ({z^2})
Clearly, option (A) is the only correct option.

Note: i=−1i = \sqrt { - 1} and i2=−1{i^2} = - 1.
The real number aa is called the real part of the complex number a+iba + ib; the real number bbis called its imaginary part. To emphasize, the imaginary part does not include a factor ii; that is, the imaginary part is bb, not ibib.