IIT JEE , AIEEE Forum - Mathematics , Algebra

shinee

1)e⁴ⁿ⁺² =-1
2)sinh^_ x = -isin^_ (ix)
3)cosh^_ x = -icos^_ (x)
4)tanh^_ x =-itan^_ (ix)
5)z =rcis(theta) then logz = logr +i(theta)
6)cos(theta) +isin(theta) = e^1(theta)
7)logi = i90⁰
8)iⁱ = e^-(4n+1)pi/2
9)sin(logiⁱ )=-1
10)if (iⁱ )^I=cos(theta) +isin(theta), then theta=(-4n+1)pi/2
11)log (1+i)= 1/2log2 +i(pi)/4

elastiboysai

e^4n+2=-1!!!!
i think u meant i^4n+2=-1
i^4n*i^2
i^4n=1,i^2=-1
=1*-1.
and culd u be more clear abt the remaining 2-- 4

5 qn is wrong again
shud be
z=re^i

logz= logre^i

=logr+loge^i

=logr+i

elastiboysai

And as for the proof of the Eulers formula here u go

using series expansions

$egin{align} i^0 &{}= 1, quad & i^1 &{}= i, quad & i^2 &{}= -1, quad & i^3 &{}= -i, i^4 &={} 1, quad & i^5 &={} i, quad & i^6 &{}= -1, quad & i^7 &{}= -i, end{align}$

and so on. The functions e^x, cos(x) and sin(x) (assuming x 2 b real ofc) can be expressed using their Taylor expansions around zero:

$egin{align} e^x &{}= 1 + x + rac{x^2}{2!} + rac{x^3}{3!} + cdots cos x &{}= 1 - rac{x^2}{2!} + rac{x^4}{4!} - rac{x^6}{6!} + cdots sin x &{}= x - rac{x^3}{3!} + rac{x^5}{5!} - rac{x^7}{7!} + cdots end{align}$

For complex z we define each of these functions by the above series, replacing x with z. This is possible because the radius of convergence of each series is infinite. We then find that

$egin{align} e^{iz} &{}= 1 + iz + rac{(iz)^2}{2!} + rac{(iz)^3}{3!} + rac{(iz)^4}{4!} + rac{(iz)^5}{5!} + rac{(iz)^6}{6!} + rac{(iz)^7}{7!} + rac{(iz)^8}{8!} + cdots &{}= 1 + iz - rac{z^2}{2!} - rac{iz^3}{3!} + rac{z^4}{4!} + rac{iz^5}{5!} - rac{z^6}{6!} - rac{iz^7}{7!} + rac{z^8}{8!} + cdots &{}= left( 1 - rac{z^2}{2!} + rac{z^4}{4!} - rac{z^6}{6!} + rac{z^8}{8!} - cdots ight) + ileft( z - rac{z^3}{3!} + rac{z^5}{5!} - rac{z^7}{7!} + cdots ight) &{}= cos (z) + isin (z) end{align}$

The rearrangement of terms is justified because each series is convergent. Taking z = x to be a real number gives the original identity as Euler discovered it.

elastiboysai

and for sinhy n coshy

${\sin(iy)\over i }= {e^{y} - e^{-y} \over 2} = \sinh(y).$
it is defined like dis.

$\cos(iy) = {e^{-y} + e^{y} \over 2} = \cosh(y)$

frm dis u can get 2,3,4
3 is wrong.

elastiboysai

$i=e^{i\pi\over2}$
from dis all oder things shud follow,
$logi={i\pi\over2}$

$i^i=e^{-\pi\over2} \\same as\\ \\e^{-\pi\cdot(4n+1)\over2}$

$\sin(log i^i)=-sin\frac{(4n+1)\pi}{2}=-1}$

elastiboysai

another method would be
$f(x) = \frac{\cos x + i\sin x}{e^{ix}}$
differentiate this function
its derivative is zero.
so f(x) is a constant function.but f(0)=1,
so f(x)=1.
so u have Eulers formula proved.
This is the simplest approach i feel.

spideyunlimited

7)

Euler's Equation states that

e^ i(pi) + 1 = 0

Thus,

e^ i(pi) = -1

So

i(pi) = ln(-1)

divide by 2:
i.pi/2 = 1/2 ln(-1)
ln ( -1)^1/2 = i . pi/2
ln i = i.90 degrees

ayshwarya

v know dat

iteta
e = Cisteta

substituting teta =90 v wud get

i90
e =i

i 90 = logi hence proved

ayshwarya

4 d last 1 its actually

1/2 log2 +ipie/4 => ipie/4 =log(1+i/rt2)

=> log rt2 +log(1+i/rt2)

=> log(1+i)

hence proved its easy

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