11Ans 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
11one more dbt
if |z1|=|z2|=|z3| and z1 z2 z3 are the vertices of an equilateral triangle such that z1+z2+z3=0 prove that this triangle is inscribed in a circle of unit radius
1i 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
11
Aditya Balasubramanyam
·2009-12-21 05:20:47
that is what u have 2 prove ... |z|=1 [22]
11
Aditya Balasubramanyam
·2009-12-21 07:58:55
ya checked ...... it s from arihant
11
Aditya Balasubramanyam
·2009-12-21 07:59:22
ya checked ...... it s from arihant and i 've solved it also...
somebody pls try q2
