If `x` is real and `(1-ix)/(1+ix)=m + i n,` show that `m^2 + n^2=1.`

If 1/ (m+i n) - (x-iy)/(x+iy) =0, where x,y,m,n are real and x+iy!=0 and m+i n!=0 , prove that m^2+n^2=1 .

(ix) ((1) / (2) + i2) ^ (3)

If m!=n and (m+n)^(-1)(m^(-1)+n^(-1))=m^xn^y then show that x+y+2=0

Show that a real value of x will satisfy the equation (1-ix)/(1+ix)=a-ib if a^(2)+b^(2)=1, wherea,breal.

The relation between the real numbers a and b ,which satisfy the equation (1-ix)/(1+ix)=a-ib, for some real value of x, is

IfI_(m , n)=int_0^(pi/2)sin^m xcos^n xdx , Then show that I_(m , n)=(m-1)/(m+n)I_m-2n(m ,n in N) Hence, prove that I_(m , n)=f(x)={((n-1)(n-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))pi/4w h e nbot hma n dna r ee v e n((m-1)(m-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))

Show that a real value of x will satisfy the equation (1-ix)/(1+ix)=a-ibquad if a^(2)+b^(2)=1 ,when a,b are real.

If I, m and n are real numbers such that l^2 + m^2 + n^2 = 0 , then show that [[1+l^2,lm,ln],[lm,1+m^2,mn],[ln,mn,1+n^2]]=1

" *.i).i) If "n" is an integer then show that "(1+i)^(2n)+(1-i)^(2n)=2^(n+1)cos(n pi)/(2)

The number of real negavitve terms in the binomial expansion of (1 + ix)^(4n-2) , n in N , n gt 0 , I = sqrt(-1) , is