Non-really Complex!

Let $z=a+bi$ , where $i=\sqrt{-1}$ is the imaginary unit and $a$ and $b$ are real numbers. $\begin{array}{rcl} z^2-26z&=&(a+bi)^2-26(a+bi)\\ &=& a^2+2abi+(bi)^2-26a-26bi\\ &=& a^2+2abi+b^2i^2-26a-26bi\\ &=& a^2+2abi-b^2-26a-26bi\\ &=&(a^2-b^2-26a)+(2ab-26b)i \end{array}$ Since $z^2-26z$ is a real number, its imaginary part is $0$ . $\begin{array}{rcl} (2ab-26b)i&=&0\\ 2ab-26b&=&0\\ 2a-26&=&0\\ 2a&=&26\\ a&=&13 \end{array}$ Therefore: $\begin{array}{rcl} z+\bar z&=&(a+bi)+(a-bi)\\ &=&(13+bi)+(13-bi)\\ &=&13+bi+13-bi\\ &=&13+13+bi-bi\\ &=&13+13\\ &=&26 \end{array}$

put z= a + ib........ so, (z)^2 = (a)^2 - (b)^2 + (2i)ab -26a -(26i)b = X +(0)i .......... this implies 2iab-26ib =0..... therefore either b=0 ( which is not possible since 'z' is bound to be a complex no. ) or a=13........ we take up a = 13 so 'z' + 'z' (conjugate)= a + ib + ( a - ib ) = 2a =26