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Nerd2567

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**1. Given, x + yi = tan^-1 ((exp(a + bi)). Prove that tan(2x) = -cos(b) / sinh(a)**

## Homework Equations

I have derived.tan(x + yi) = i*tan(x)*tanh(y) / 1 - i*tan(x)*tanh(y)

tan(2x) = 2tanx / 1 - tan^2 (x)

Exp(a+bi) = exp(a) *(cos(b) + i*sin(b))[/B]

**3. My attempt:**

By definition sinha = (exp(a) - exp (-a)) / 2

Let x + yi = z

tanz = exp(a + bi) = exp(a) * exp(bi)

hence; exp(a) = tan(z) / exp(bi)

exp(-a) = exp(bi) /tan(z)

therefore; exp(a) - exp(-a) = (tan(z) / exp(bi) ) * ( exp(bi) / tan(z) )

simplifying; exp(a) - exp(-a) = (tan^2(z) - exp(2bi)) / (exp(bi) * tan(z))

hence, sinh (a) = (

By definition sinha = (exp(a) - exp (-a)) / 2

Let x + yi = z

tanz = exp(a + bi) = exp(a) * exp(bi)

hence; exp(a) = tan(z) / exp(bi)

exp(-a) = exp(bi) /tan(z)

therefore; exp(a) - exp(-a) = (tan(z) / exp(bi) ) * ( exp(bi) / tan(z) )

simplifying; exp(a) - exp(-a) = (tan^2(z) - exp(2bi)) / (exp(bi) * tan(z))

hence, sinh (a) = (

**exp(a) - exp(-a))/ 2 =****(tan^2(z) - exp(2bi)) / 2(exp(bi) * tan(z))**

Now am going to find -cos(b) --------

tanz = exp(a) * (cos(b) + isin(b))

cos(b) = (tan(z) / exp(a)) - isin(b)

-cos(b) = isin(b) - (tan(z) / exp(a))

simplifying;

-cos(b) = ( iexp(a)*sin(b) - tan(z) ) / exp(a)

therefore;

-cos(b) / sinh (a) =((( 2iexp(a + bi) * sin(b) tan(z) - 2exp(bi) * tan^2 (z) ))) / (( exp(a) tan^2(z) - exp(a +2bi) ))

this is pretty much my best attempt...:'( I must be missing theory please help...I will really appreciate it!

Now am going to find -cos(b) --------

tanz = exp(a) * (cos(b) + isin(b))

cos(b) = (tan(z) / exp(a)) - isin(b)

-cos(b) = isin(b) - (tan(z) / exp(a))

simplifying;

-cos(b) = ( iexp(a)*sin(b) - tan(z) ) / exp(a)

therefore;

-cos(b) / sinh (a) =((( 2iexp(a + bi) * sin(b) tan(z) - 2exp(bi) * tan^2 (z) ))) / (( exp(a) tan^2(z) - exp(a +2bi) ))

this is pretty much my best attempt...:'( I must be missing theory please help...I will really appreciate it!