IIT JEE , AIEEE Forum - Mathematics , Algebra

aks_0123456

Q1 Real part of e^eiota@ (*iota@ is a power to second e)

Q2 if 3+i = (3 - i)z' ,then z=? (* z' is conjugate of z)

Q3 prove that z₁ = x₁ +iy₁ ,z₂ = x₂ + iy₂ are perpendicular iff z₁z₂' +z₁'z₂ =0

(* z' is conjugate of z)

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sboosy

Let j be the complement of z where z= x+iy . so j=x-iy
3+i = (3-i) j
3+i = (3-i)*(x-iy)
3+i = 3x-y + i (-x-3y)
comparing real and imaginary parts
and solving the two eqns we get
z= 4/5 + i (-3/5)

hsbhatt

Qn 3: I think you meant, the lines joining z₁and z₂ are perpendicular.

Consider triangle OZ₁Z₂right angled at Z₁.

By Pythagoras theorem (Z₁Z₂)² = OZ₁²+ OZ₂²

Now OZ₁ = |z₁|; OZ₂ = |z₂| and Z₁Z₂ = |z₁-z₂|

Hence |z₁|²+|z₂|²= |z₁-z₂|² = (z₁-z₂) (z'₁-z_2^') = |z₁|²+|z₂|²- (z₁z₂' +z₁'z₂)

Hence z₁z₂' +z₁'z₂ = 0.

neeraj_agarwal_1990

Q1

e^(cos@+isin@) = e^cos@ . e^isin@

write e^isin@ as cos(sin@) + isin(sin@)

real part is e^cos@ . [cos(sin@)]

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