Question

Express 101010100₂ to denary

Answer 1

Answer:

Real part = \dfrac{2\cos x\cos hy}{\cos 2x+\cosh 2y}

cos2x+cosh2y

2cosxcoshy

and Imaginary part = \dfrac{2\sin x\sin hy}{\cos 2x+\cosh 2y}

cos2x+cosh2y

2sinxsinhy

Step-by-step explanation:

We have,

\sec(x+iy)sec(x+iy)

To separate the real and imaginary part of \sec(x+iy)sec(x+iy) = ?

∴ \sec(x+iy)sec(x+iy)

=\dfrac{1}{\cos(x+iy)}=

cos(x+iy)

1

Rationalising numerator and denominator, we get

=\dfrac{1}{2\cos(x+iy)}\times \dfrac{2\cos(x-iy)}{\cos(x-iy)}=

2cos(x+iy)

1

×

cos(x−iy)

2cos(x−iy)

=\dfrac{2(\cos x\cos iy+\sin x\sin iy)}{\cos 2x+\cos 2iy}=

cos2x+cos2iy

2(cosxcosiy+sinxsiniy)

Using the trigonometric identity,

\cos (A-B)=\cos A\cos B+\sin A\sin Bcos(A−B)=cosAcosB+sinAsinB

=\dfrac{2(\cos x\cos hy+i\sin x\sin hy)}{\cos 2x+\cosh 2y}