Question

Express the following in the form of a+ib

i²⁰¹⁹

Answer 1

Answer:

$i^{2019}=-i$

Step-by-step explanation:

$i^{2019}$

$\gray{\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c}$

$\gray{i^{2019}=i^{2018}i}$

$=i^{2018}i$

$\gray{\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=\left(a^b\right)^c}$

$\gray{i^{2018}=\left(i^2\right)^{1009}}$

$=i\left(i^2\right)^{1009}$

$\gray{\mathrm{Apply\:imaginary\:number\:rule}:\quad \:i^2=-1}$

$=\left(-1\right)^{1009}i$

$\gray{\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=-a^n,\:\mathrm{if\:}n\mathrm{\:is\:odd}}$

$\gray{\left(-1\right)^{1009}=-1^{1009}}$

$=-1^{1009}i$

$\gray{1^{1009}=1}$

$=-1i$

$\gray{\mathrm{Multiply:}\:1i=i}$

$=-i$