The-imaginary-part-of-1-2-1-2-i-10-is-

Question Number 58402 by rahul 19 last updated on 22/Apr/19

T h e i m a g i n a r y p a r t o f {(\frac{1}{2} + \frac{1}{2} i)}^{10} i s ?

Commented by maxmathsup by imad last updated on 23/Apr/19

l e t Z = {(\frac{1}{2} + \frac{1}{2} i)}^{10} \Rightarrow Z = \frac{1}{2^{10}} {(1 + i)}^{10} = \frac{1}{2^{10}} {(\sqrt{2})}^{10} {(e^{\frac{i π}{4}})}^{10} = \frac{1}{2^{5}} e^{i \frac{5 π}{2}}

= \frac{1}{2^{5}} e^{\frac{i π}{2}} = \frac{1}{2^{5}} {c o s (\frac{π}{2}) + i s i n (\frac{π}{2})} = \frac{i}{2^{5}} \Rightarrow I m (Z) = \frac{1}{2^{5}}

Commented by rahul 19 last updated on 23/Apr/19

t h a n k s s i r!

Answered by tanmay last updated on 23/Apr/19

\frac{1}{2^{10}} ({(1 + i)}^{10}

= \frac{1}{2^{10}} {[\sqrt{2} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i)]}^{10}

\frac{1}{2^{10}} \times 2^{5} {[c o s \frac{π}{4} + i s i n \frac{π}{4}]}^{10}

= \frac{1}{2^{5}} [c o s (\frac{5 π}{2}) + i s i n (\frac{5 π}{2})]

= \frac{1}{2^{5}} [0 + i]

= i \times \frac{1}{2^{5}}

= r e a l p a r t = 0

i m a g i n a r y = 1 \times \frac{1}{2^{5}}

Commented by rahul 19 last updated on 22/Apr/19

A n s \to \frac{1}{2^{5}} [∵ \frac{1}{2^{10}} w i l l b e t a k e n a s c o m m o n

i n 1 s t s t e p] .

t h a n k s s i r \dots .

g o o d n i g h t & s w e e t d r e a m s :)

Commented by tanmay last updated on 23/Apr/19

t h a n k y o u r a h u l \dots s h u t t e r o f w i n d o w o f m i n d

(e y e s) g o t f a t i g u e d \dots s o \dots .

Answered by MJS last updated on 23/Apr/19

{(\frac{\sqrt{2}}{2} e^{i \frac{π}{4}})}^{10} = \frac{1}{32} e^{i \frac{5 π}{2}} = \frac{1}{32} i \Rightarrow answer is \frac{1}{32}

Commented by rahul 19 last updated on 23/Apr/19

t h a n k s s i r!