How do you simplify 5(cospi+isinpi)*2(cos((3pi)/4)+isin((3pi)/4))5(cosπ+isinπ)2(cos(3π4)+isin(3π4)) and express the result in rectangular form?

1 Answer
May 17, 2017

5sqrt2(1-i).52(1i).

Explanation:

{5(cospi+isinpi)}{2(cos(3pi/4)+isin(3pi/4))}{5(cosπ+isinπ)}{2(cos(3π4)+isin(3π4))}

={5(-1+0i)}{2(-cos(pi/4)+isin(pi/4))}={5(1+0i)}{2(cos(π4)+isin(π4))}

=-5{2(-1/sqrt2+1/sqrt2*i)}=5{2(12+12i)}

=5sqrt2(1-i),=52(1i),

Otherwise, knowing that,

r(cosalpha+isinalpha)*R(cosbeta+isinbeta)=(rR)(cos(alpha+beta)+isin(alpha+beta)),r(cosα+isinα)R(cosβ+isinβ)=(rR)(cos(α+β)+isin(α+β)),

we have, the Reqd. Value

=(5xx2){cos(pi+3pi/4)+isin(pi+3pi/4)}=(5×2){cos(π+3π4)+isin(π+3π4)}

=10{cos(2pi-pi/4)+isin(2pi-pi/4)}=10{cos(2ππ4)+isin(2ππ4)}

=10{cospi/4-isinpi/4}=10{cosπ4isinπ4}

=10(1/sqrt2-i/sqrt2)=10(12i2)

=5sqrt2(1-i),=52(1i), as derived earlier!

Enjoy Maths.!