Question

for any complex number z prove that Re(z) = z+z'/2 and I'm(z) =z-z'/2i
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Answer 1

$\underline{\textbf{To prove:}}$

$\textsf{For any complex number z,}$

$\mathsf{Re(z)=\dfrac{z+z'}{2}\;\;\&\;\;Im(z)=\dfrac{z-z'}{2i}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Let\;z=a+i\,b}$

$\mathsf{Here,\;Re(z)=a\;\7\;Im(z)=b}$

$\mathsf{Then,\;z'=a-i\,b}$

$\mathsf{z+z'=(a+i\,b)+(a-i\,b)}$

$\mathsf{z+z'=2\;a}$

$\mathsf{z+z'=2\;Re(z)}$

$\implies\boxed{\mathsf{Re(z)=\dfrac{z+z'}{2}}}$

$\mathsf{z-z'=(a+i\,b)-(a-i\,b)}$

$\mathsf{z-z'=a+i\,b-a+i\,b}$

$\mathsf{z-z'=2i\,b}$

$\mathsf{z-z'=2i\,Im(z)}$

$\implies\boxed{\mathsf{Im(z)=\dfrac{z-z'}{2i}}}$

$\underline{\textbf{Find more:}}$

Write (a+ib/a-ib)^2-(a-ib/a+ib)^2 in the form x +iy please reply me this

https://brainly.in/question/18153048

Answer 2

Answer:

Let z=a+ib

then,z'=a-ib

Re(z)=a

Im (z)=b

Z+Z'=a+ib+a-ib

or,Z+Z'=2a

or,Z+Z'=2Re(z)

or,Re(z)=z+z'/2

Now,

Z-Z'=a+ib-a+ib

or,Z-Z'=2ib

or,Z-Z'=2i I'm(z)

or,Im(z)=Z-Z'/2i

hence ,proved