Question

Evaluate: ${\left(1+i\right)}^6+{\left(1-i\right)}^3$

Answer 1

Evaluating the given exp[ression:

Consider ${\left(1+i\right)}^6+{\left(1-i\right)}^3$

but ${\left(1+i\right)}^6={\left({\left(1+i\right)}^2\right)}^3$

We know that ${\left(a+b\right)}^2=a^2+b^2+2ab$

$$\begin{gathered} \Rightarrow {\left(1+i\right)}^2=1^2+i^2+2\left(1\right)\left(i\right) \\ =1+\left(-1\right)+2i\left[i^2=-1\right] \\ =1-1+2i \\ =2i \\ \therefore {\left({\left(1+i\right)}^2\right)}^3={\left(2i\right)}^3 \\ =8i^3\left[i^3=-i\right] \\ =-8i \end{gathered}$$

Also,

$$\begin{gathered} {\left(1-i\right)}^3=1-i^3-3i+3i^2 \\ =1-\left(-i\right)-3i+3\left(-1\right) \\ =1+i-3i-3 \\ =-2-2i \end{gathered}$$

$$\begin{gathered} \therefore {\left(1+i\right)}^6+{\left(1-i\right)}^3=-8i+\left(-2-2i\right) \\ =-8i-2-2i \\ =-2-10i \end{gathered}$$

Hence,${\left(1+i\right)}^6+{\left(1-i\right)}^3=-2-10i$.