Question

A value of n for which (1+i ) Power n - (l-i) Power n =zero

Answer 1

$\huge\boxed{\underline{\underline{\green{Solution}}}}$

COMPLEX NUMBER

A Complex Number is number which can be expressed in the form $a + ib$

where a & b are a real numbers

$i =$Complex Number satisfying ${i}^{2} = - 1$

$\displaystyle \: \hookrightarrow \:$TO DETERMINE

A value of n such that $\displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n} = 0$

$\displaystyle \: \hookrightarrow \:$CALCULATION

Taking n = 1

$\displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n}$

$= \displaystyle \: {( 1 + i)}^{1} + {( 1 - i)}^{1}$

$= \displaystyle \: ( 1 + i) + ( 1 - i) = 2 \ne \: 0$

Taking n = 2

$\displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n}$

$= \displaystyle \: {( 1 + i)}^{2} + {( 1 - i)}^{2}$

$= \displaystyle \: ( 1 + 2i + {i}^{2} ) + ( 1 - 2i + {i}^{2} )$

$\displaystyle \: = 1 + 2i - 1 + 1 - 2i - 1$

$= 0$

$\displaystyle \: \hookrightarrow \:$RESULT

Hence 2 is least value of n such that

$\displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n} = 0$

Answer 2

Answer:

COMPLEX NUMBER

A Complex Number is number whichcan be expressed in the form a + iba+ib

where a & b are a real numbers

i =i= Complex Number satisfying {i}^{2} = - 1i2=−1

\displaystyle \: \hookrightarrow \:↪ TO DETERMINE

A value of n such that \displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n} = 0(1+i)n+(1−i)n=0

\displaystyle \: \hookrightarrow \:↪ CALCULATION

Taking n = 1

\displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n}(1+i)n+(1−i)n

= \displaystyle \: {( 1 + i)}^{1} + {( 1 - i)}^{1}=(1+i)1+(1−i)1

= \displaystyle \: ( 1 + i) + ( 1 - i) = 2 \ne \: 0=(1+i)+(1−i)=2=0

Taking n = 2

\displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n}(1+i)n+(1−i)n

= \displaystyle \: {( 1 + i)}^{2} + {( 1 - i)}^{2}=(1+i)2+(1−i)2

= \displaystyle \: ( 1 + 2i + {i}^{2} ) + ( 1 - 2i + {i}^{2} )=(1+2i+i2)+(1−2i+i2)

\displaystyle \: = 1 + 2i - 1 + 1 - 2i - 1=1+2i−1+1−2i−1

= 0=0

\displaystyle \: \hookrightarrow \:↪ RESULT

Hence 2 is least value of n such that

\displaystyle \: {( 1 + i)}^{n} + {( 1 - i)}^{n} = 0(1+i)n+(1−i)n=0