Question

Find the multiplicative reciprocal of 1of3/6​

Answer 1

Answer )-

Let the multiplicative inverse of 5−i be a+bi

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b=

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 26

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a=

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 16

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165 Therefore, multiplicative inverse is

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165 Therefore, multiplicative inverse is 26

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165 Therefore, multiplicative inverse is 26(5+i)

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165 Therefore, multiplicative inverse is 26(5+i)

Let the multiplicative inverse of 5−i be a+bi Therefore we have (5−i)(a+bi)=1⇒5a+b+i(5b−a)=1 ⇒5a+b=1 and 5b−a=0⇒a=5b Substitute this in 5a+b=1, we get b= 261 and a= 165 Therefore, multiplicative inverse is 26(5+i) .