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Question: If \(z\) is the complex number such that \(\operatorname{Re} (z) = \operatorname{Im} (z)\) then A...

If zz is the complex number such that Re⁡(z)=Im⁡(z)\operatorname{Re} (z) = \operatorname{Im} (z) then
A. Re⁡(z2)=0\operatorname{Re} ({z^2}) = 0
B. Im⁡(z2)=0\operatorname{Im} ({z^2}) = 0
C. Re⁡(z2)=Im⁡(z2)\operatorname{Re} ({z^2}) = \operatorname{Im} ({z^2})
D. Re⁡(z2)=−Im⁡(z2)\operatorname{Re} ({z^2}) = - \operatorname{Im} ({z^2})

Explanation

Solution

Let z=a+ibz = a + ib so Re⁡(z)=a,Im⁡(z)=b\operatorname{Re} (z) = a,\operatorname{Im} (z) = b and it is also given that a=ba = b so z=a+ia=a(1+i)z = a + ia = a(1 + i) and now we can find z2{z^2}.

Complete step by step solution:
Here we are given that zz is the complex number and we know that complex number is formed by the real part denoted by Re⁡(z)\operatorname{Re} (z) and the imaginary part which is given by Im⁡(z)\operatorname{Im} (z) and here real part is the constant term while the imaginary part is the coefficient of ii which is iota and is given as i=−1i = \sqrt { - 1} . So let us assume z=a+ibz = a + ib
So here aa is the real part and bb is the imaginary part which means that Re⁡(z)=a,Im⁡(z)=b\operatorname{Re} (z) = a,\operatorname{Im} (z) = b
And we are also given that imaginary and the real part of the complex function are equal which means that a=ba = b
So as we know that a=ba = b so we can replace bb with aa
So
z=a+ibz = a + ib
z=a+ia=a(1+i)z = a + ia = a(1 + i)
Now we need to find the z2{z^2}
So upon squaring both the sides we will get that
z2=a2(1+i)2{z^2} = {a^2}{(1 + i)^2}
Here we also know that (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
So applying this formula in the above equation we will get that
z2=a2(1+i2+2i){z^2} = {a^2}(1 + {i^2} + 2i)
=a2(1+(−1)+2i)= {a^2}(1 + ( - 1) + 2i)
=a2(1−1+2i)= {a^2}(1 - 1 + 2i)
=2a2i= 2{a^2}i
So we know that z2{z^2} is another complex number whose real part is 00 and imaginary part is 2a22{a^2}

So we can say that Re⁡(z2)=0\operatorname{Re} ({z^2}) = 0 when it is given that Re⁡(z)=Im⁡(z)\operatorname{Re} (z) = \operatorname{Im} (z)

Note:
As we know that ω,ω2\omega ,{\omega ^2} also represent the complex number which are given by ω=−1+3i2\omega = \dfrac{{ - 1 + \sqrt 3 i}}{2}
And ω2=−1−3i2{\omega ^2} = \dfrac{{ - 1 - \sqrt 3 i}}{2} and we know that ω2=1{\omega ^2} = 1 and hence we can say that ω2+ω+1=0{\omega ^2} + \omega + 1 = 0