Math, asked by Thatsomeone, 6 months ago

\tt For\:positive\:integer\:{n}_{1},{n}_{2} \:the\:value\:of\:expression \\ \\ \tt {(1+i)}^{{n}_{1}} + {(1+{i}^{3})}^{{n}_{1}} + {(1+{i}^{5})}^{{n}_{2}} + {(1+{i}^{7})}^{{n}_{2}} \\ \\ \tt Where\:i = \sqrt{-1} \:is\:a\:real\:number\:if\:and\:only\:if \\ \\ \tt a. {n}_{1} = {n}_{2} + 1 \\ \tt b. {n}_{1} = {n}_{2} - 1 \\ \tt {n}_{1} = {n}_{2} \\ \tt {n}_{1}>0 , {n}_{2}>0

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Answered by AVENGERS789456
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Answered by Anonymous
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\tt For\:positive\:integer\:{n}_{1},{n}_{2} \:the\:value\:of\:expression \\ \\ \tt {(1+i)}^{{n}_{1}} + {(1+{i}^{3})}^{{n}_{1}} + {(1+{i}^{5})}^{{n}_{2}} + {(1+{i}^{7})}^{{n}_{2}} \\ \\ \tt Where\:i = \sqrt{-1} \:is\:a\:real\:number\:if\:and\:only\:if \\ \\ \tt a. {n}_{1} = {n}_{2} + 1 \\ \tt b. {n}_{1} = {n}_{2} - 1 \\ \tt {n}_{1} = {n}_{2} \\ \tt {n}_{1}>0 , {n}_{2}>0

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