====== Question 5, Exercise 1.2 ======
Solutions of Question 5 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 
====Question 5====
If $z_1$ and $z_2$ are two any complex numbers then prove that $|z_1+z_2|^2-|z_1-z_2|^2=4Re(z_1)Re(z_2)$FIXME

**Solution.** 
Suppose \begin{align}z_1&=x_1+iy_1 \text{ and } z_2&=x_2+iy_2\end{align}
Now
\begin{align}z_1+z_2&=x_1+iy_1+x_2+iy_2\\
 &=x_1+x_2+i(y_1+y_2)\\
|z_1+z_2|^2&=(x_1+x_2)^2+(y_1+y_2)^2\\
 &=x^2_1+x^2_2+2x_1x_2+y^2_1+y^2_2+2y_1y_2 ... (1)\end{align}
\begin{align}
z_1-z_2&=x_1+iy_1-(x_2+iy_2)\\
&=x_1-x_2+i(y_1-y_2)\\
|z_1-z_2|^2&=(x_1-x_2)^2+(y_1-y_2)^2\\
 &=x^2_1+x^2_2-2x_1x_2+y^2_1+y^2_2-2y_1y_2 ... (2)\end{align}
Now from (1) and (2)
\begin{align}&|z_1+z_2|^2-|z_1-z_2|^2 \\
=&(x^2_1+x^2_2+2x_1x_2+y^2_1+y^2_2+2y_1y_2) \\&-(x^2_1+x^2_2-2x_1x_2+y^2_1+y^2_2-2y_1y_2)\\
=&2x_1x_2+2y_1y_2+2x_1x_2+2y_1y_2\\
=&4x_1x_2+4y_1y_2 \\
=&4Re(z_1)Re(z_2)+4Im(z_1)Im(z_2).\end{align}
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