Chapter P.6, Problem 56E

Solution

To prove the complex conjugate of a sum of two complex number is the sum of their complex conjugate.

Concept used:

• Product of powers property, $a^n\cdot a^m=a^{n+m}$
• FOIL to multiply binomials, $\left(a+b\right)\left(c+d\right)=a\cdot c+a\cdot d+b\cdot c+b\cdot d$
• Properties of complex number: $\sqrt{-1}=i,i^2=-1,i^3=-i$
• The complex conjugate of $a+bi$ is $a-bi$ .
• Difference of squares formula, $\left(a+b\right)\left(a-b\right)=a^2-b^2$

Calculation:

In order to prove the complex conjugate of a sum of two complex number is the sum of their complex conjugate, the two complex number are $a+biandc+di$ first solving these two complex number by adding both complex number and then adding their conjugates as shown below:

  $\begin{array}{l}\left(a+bi\right)+\left(c+di\right)\\ a+bi+c+di\\ \left(a+c\right)+\left(b+d\right)i\end{array}$

This is the sum of the number , now take the conjugate of this :

  $\left(a+c\right)-\left(b+d\right)i$

Now, sum of the conjugate as shown below:

  $\begin{array}{l}\left(a-bi\right)+\left(c-di\right)\\ a-bi+c-di\\ a+c-\left(b+d\right)i\end{array}$

So, the two results from both parts are the same ,hence proved.