Math, asked by clarafanai17, 10 months ago

6i^50+5i^33-2i^15+6i^48

Answers

Answered by AbhijithPrakash
6

Answer:

6i^{50}+5i^{33}-2i^{15}+6i^{48}=7i

Step-by-step explanation:

6i^{50}+5i^{33}-2i^{15}+6i^{48}

\mathrm{Let's\:first\:take:\:}6i^{50}

6i^{50}

\mathrm{To\:find\:the\:value\:of\:}i^{50}

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

i^{50}=\left(i^2\right)^{25}

=\left(i^2\right)^{25}

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

=\left(-1\right)^{25}

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

\left(-1\right)^{25}=-1^{25}

=-1

\mathrm{Plug\:in\:the\:value\:of\:}i^{50}\mathrm{\:in\:}6i^{50}

=6\left(-1\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

=-6\cdot \:1

\mathrm{Multiply\:the\:numbers:}\:6\cdot \:1=6

=-6

=-6+5i^{33}-2i^{15}+6i^{48}

\mathrm{Similarly;}

5i^{33}

i^{33}

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

i^{33}=i^{32}i

=i^{32}i

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

i^{32}=\left(i^2\right)^{16}

=i\left(i^2\right)^{16}

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

=\left(-1\right)^{16}i

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

\left(-1\right)^{16}=1^{16}

=1^{16}i

1^{16}=1

=1i

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

=i

=5i

2i^{15}

i^{15}=-i

=2\left(-i\right)

6i^{48}

i^{48}=1

=6\cdot \:1

\mathrm{Multiply\:the\:numbers:}\:6\cdot \:1=6

=6

\mathrm{Now,\:plug\:in\:the\:results\:back\:in;}

=-6+5i-\left(-2i\right)+6

\mathrm{Apply\:rule}\:-\left(-a\right)=a

=-6+5i+2i+6

\mathrm{Add\:similar\:elements:}\:5i+2i=7i

=-6+7i+6

\mathrm{As,\:}-6+6=0

=7i

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