Optionals IAS Mains Mathematics: Questions 87 - 92 of 283

Given

and

And

is harmonic function.

Now

[By C. R. Equation ]

Its solution is

is harmonic conjugate of

is analytic function.

Now Let

lemma: Let be a vector space over a field Then is a subspace of its and

Proof: - Let itself is a vector space over .

Then itself is a vector space over

for all and .

Conversely.

Let and .

we get

is an additive abelian group.

Also taking we have

is closed under scalar multiplication. The remaining axioms for scalar multiplication also hold for…

We know that Basis of is

First we shall find Basis for Image of

is Basis of

generates Image of .

Now is defined as

Here

Generate Image of

To find basis of Range , we have to find the L. I Vectors from for this consider the Matrix whose rows are generator of and reduce it to Echelon form.

i.e..

Operate

Operate

from L. I set of vectors which genera…

Clearly the function f (z) is not analytic at z = 1 and z = 2.

When

When

(Since Binomial expansion is valid when )

(iii) when|Z|> 2

Binomial expansion is valid when

Given,

Now,

and

Again

Let along we have

And if let along we have

Hence does not exist.

So, is not analytic at origin.

Now,

Operate

Operate

Operate

So are solutions of

Now,

Operate

Operate

So, are solution of .