36is it something like ....
√4 - w2
Let z and w be two complex numbers such that |z|=|w|=1
and |z+iw|=|z-iw|=2. then z=
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3 Answers
21I think that the soln. does not exist.
|z|=|w|=|iw|=1
Now to get mod value as 2 the argument of z,iw & that of z,(-iw) should be same.However this means that arg(wi)=arg(-wi) which is impossible.
Hence no soln.
Here the sum and diff. of 2 unit vectors have same mod value hence they must be inclined at 90.The mod value at this pos. will be √2(and not 2 for soln. to exist).The value of z then will be ±w.
262|z+iw| ≤|z| +|iw| (equality occurs when both are parallel ie. have same argument)
now |z|=|iw|=1
thus |z+iw|≤2 but |z+iw|=2
hence z = iw
but |z-iw|=2 which is not possible from the above relation ,
hence no such z exists.
