Question

express as x+iy , i^21-i^66+3i^48​

Answer 1

Answer:

The given complex number in the form of x + iy is 4 + i.

Step-by-step-explanation:

We have given a complex number.

We have to express it in standard form x + iy.

The given complex number is

$\displaystyle{\sf\:i^{21}\:-\:i^{66}\:+\:3i^{48}}$

Let this complex number be z.

$\displaystyle{\therefore\:\sf\:z\:=\:i^{21}\:-\:i^{66}\:+\:3i^{48}}$

$\displaystyle{\implies\sf\:z\:=\:i^{20\:+\:1}\:-\:i^{64\:+\:2}\:+\:3\:i^{44\:+\:4}}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:a^{m\:+\:n}\:=\:a^m\:\times\:a^n\:}}}$

$\displaystyle{\implies\sf\:z\:=\:i^{20}\:\times\:i\:-\:i^{64}\:\times\:i^2\:+\:3\:i^{44}\:\times\:i^4}$

$\displaystyle{\implies\sf\:z\:=\:i^{4\:\times\:5}\:\times\:i\:-\:i^{4\:\times\:16}\:\times\:i^2\:+\:3\:i^{4\:\times\:11}\:\times\:i^4}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:a^{m\:\times\:n}\:=\:(\:a^m\:)^n\:}}}$

$\displaystyle{\implies\sf\:z\:=\:(\:i^4\:)^5\:\times\:i\:-\:(\:i^4\:)^6\:\times\:i^2\:+\:3\:(\:i^4\:)^{11}\:\times\:i^4}$

We know that,

$\displaystyle{\boxed{\green{\sf\:i^2\:=\:-\:1\:}}}$

$\displaystyle{\boxed{\purple{\sf\:i^4\:=\:1\:}}}$

$\displaystyle{\implies\sf\:z\:=\:1^5\:\times\:i\:-\:1^6\:\times\:(\:-\:1\:)\:+\:3\:\times\:1^{11}\:\times\:1}$

$\displaystyle{\implies\sf\:z\:=\:1\:\times\:i\:-\:1\:\times\:(\:-\:1\:)\:+\:3\:\times\:1}$

$\displaystyle{\implies\sf\:z\:=\:i\:+\:1\:+\:3}$

$\displaystyle{\implies\sf\:z\:=\:i\:+\:4}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:z\:=\:4\:+\:i\:}}}}$

∴ The given complex number in the form of x + iy is 4 + i.