Complex Nos

Ans 1) i ^{i i i .....∞} = a + i b
i ^{a+ i b} = a + i b

Now, a + i b = i ^{a+ i b} = e^{(a+i b ) log i} [Taking Principal values only]

= e^{(a+i b ) log (cos pie/2 + i sin pie/2)i}
= e^{(a+i b ) log (e i pie/2)}
= e^{(a+i b ) i pie/2} = e ^{- ( b pie/2) + i (a pie/2)}
= e ^{- b pie/2} . e ^{i a pie/2}

= e ^{- b pie/2} (cos a pie/2 + i a sin pie/2)
Equating real and imaginary parts, we get

a = e ^{- b pie/2} cos (a pie/2) .........................(i)
b = e ^{- b (pie/2)} sin (a pie/2) .........................(ii)

Dividing (ii) by (i) ,
tan (a pie/2) = b/a

Squaring and adding (i) and (ii), we get
a² + b² = e ^{- b pie} [cos ²( apie/2) + sin ² (a pie/2)] = e ^{- b pie}

Thus Proved

thanx tushar

i guess u cant prove it to be inscribe in a circle of unit radius unless |z1|=|z2|=|z3| =1

centriod here is origin z1+z2+z3/3=0/3=0
for equil trian circumcenter is also centriod

so circumcent is origin and if it wud hav been given |z1|=|z2|=|z3|=1 then we wud hav proved that tat circle is of unit radius

that is what u have 2 prove ... |z|=1 [22]

nope check the q again

ya checked ...... it s from arihant

ya checked ...... it s from arihant and i 've solved it also...

somebody pls try q2