Don't know the answer. But seems to be right. I don't like it too. :P
May you take me to an deeper exploration of Geometrical Concepts of Complex Nos. which would be greatly helpful for IIT-JEE?
1. If sin (log i i) = a+ib. Find a and b.
2. Find the area formed by z , -iz and z+iz. I solved it but I need the Geometrical Interpretation!!
2.
Area of OZ(iZ) = 12 |z|. |z| = |z|22
so total area = 2 (Area of OZ(iZ)) = |z|2
Please confirm if its right....
Don't know the answer. But seems to be right. I don't like it too. :P
May you take me to an deeper exploration of Geometrical Concepts of Complex Nos. which would be greatly helpful for IIT-JEE?
1 . I am going to use only principal values .
i i = e ( i π / 2 ) x i = e - π / 2
So , ln ( i i ) = - π2
Now , clearly , " a = - 1 " and " b = 0 " .
2 . Multiplying a complex number " z " with " i " simply turns the direction of " z " by a right angle in the anti - clockwise sense , i . e , the phasor " z " is rotated by 90 degrees . So , if you think in terms of vetors , the three quantities , " z" , " i z " and " z + i z " are nothing but three sides of a right angled triangle .The triangle can be scetched as -
Now ,you might as well tackle the problem you asked using this easy but handy concept .
I solved first one like this... Is it correct
sin (log i i)
= sin { log (cos π/2 + i sin π/2)i }
= sin {log( e iθ) i }
= sin { log e -Ï€/2 }
= sin -Ï€/2
= -1
Is it correct?
Also, How to draw Argand graphs?
Secondly,
If x = eiθ and √(1-c2) = nc-1, then prove that
1+ccos\theta = \frac{c}{2n} (1+nx) (1+\frac{n}{x}
-iz.
The figure of Ricky is wrong I think. But thanks for help.
Actually I drew the graph for " z" , " i z " and " z + i z " only , if you see my text .
As for Vivek ' s query ,
If , x = cos a + i sin a ;
then x + 1x = 2 cos a ...........Surely you understand
Now , given , √1 - c 2 = n c -1
or , 1 - c 2 = n 2 c 2 - 2 n c + 1
or , c 2 ( n 2 + 1 ) = 2 n c
or , n 2 + 1 = 2 n c
So , c2n ( 1 + n x ) ( 1 + nx )
= c2 n { 1 + n ( x + 1x ) + n 2 }
= c2 n ( 2 n cos a + n 2 + 1 )
= c cos a + c2 n x 2 n c
= 1 + c cos a .
Thanks.. Got that my approach was right.
\sum_{p=1}^{32}{(3p+2) \left\{\sum_{q=1}^{k}{(sin\frac{2q\Pi }{11}-icos\frac{2q\Pi }{11}} \right\}}^{p}
I solved this too.. But that k in the second Summation is a nagging one, it isn't disappearing and answer doesn't have any k.
If z1 and z2 are any two complex no. then prove that:
|z_{1}+z_{2}|\geq \frac{1}{2} (|z_{1}|+|z_{2}|)\left|\frac{z_{1}}{|z_{1}|} + \frac{z_{2}}{|z_{2}|} \right|
For the summation question, i think it will be summation of q from 1 to 10 and not 1 to k.
Same question solved here:
http://targetiit.com/iit-jee-forum/posts/complex-numers-12988.html
Wahi toh.. Main thak gaya k ko gayab karne mein.
Tab toh solve ho gaya.!! :)
Last one!