Question

find the value of and it is from lesson - complex number of 11th standard.
answer it guys ​

Answer 1

Answer with Explanation :

First, find the values of $\sf i^6,i^7,i^{11}$

$\sf (i)^6=(i^2)^3=((\sqrt{-1})^2)^3=(-1)^3=-1$

$\sf (i)^7=(i)^6(i)=(-1)(i)=-i$

$\sf (i)^{11}=(i)^{10}(i)=(i^2)^5(i)$

$=\sf ((\sqrt{-1})^2)^5(i)=(-1)^5(i)=-i$

Therefore

$\sf (3+\frac{2}{i})(i^6-i^7)(1+i^{11})$=

$\implies \sf (3+\frac{2}{i})(-1-(-i)) (1+(-i))$

$\implies \sf (3+\frac{2}{i})(-1+i)(1-i)$

$\implies \sf (-3+3i-\frac{2}{i}+2)(1-i)$

$\implies \sf (-1+3i-\frac{2}{i})(1-i)$

$\implies \sf -1+i+3i-3i^2-\frac{2}{i}+2$

$\implies \sf -1+4i+3-(\frac{2-2i}{i})$

$\implies \sf 4i + 2-(\frac{2-2i}{i})$

$\implies \sf (\frac{4i^2+2i-2+2i}{i})$

$\implies \sf (\frac{-4+4i-2}{i})$

$\implies \sf (\frac{-6+4i}{i})$

$\implies \sf (\frac{(4i-6)\times(i)}{i\times i}$

$\implies \sf (\frac{-4-6i}{-1})$

$\implies \boxed{\boxed{\boxed{\boxed{\sf 4+6i}}}}$