Can you find the product of z1 and z2? z1=4(cos40˚+ i sin40˚) z2=2(cos20˚+ i sin20˚) Give your answer in rectangular form.

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Answer 1 · 2018-05-30T07:42:08

$z_{1} z_{2} = 4 + 4 \sqrt{3} i$ $z_1z_2=4+4sqrt3i$

there are two ways of doing this

method 1

$z_{1} = r_{1} (cos θ + i sin θ)$ $z_1=r_1(costheta+isintheta)$

$z_{2} = r_{2} (cos ϕ + i sin ϕ)$ $z_2=r_2(cosphi+isinphi)$

$\Rightarrow z_{1} z_{2} = r_{1} r_{2} (cos (θ + ϕ) + i sin (θ + ϕ)) - - (1)$ $=>z_1z_2=r_1r_2(cos(theta+phi)+isin(theta+phi))--(1)$

we have

$z_{1} = 4 (cos 40 + i sin 40)$ $z_1=4(cos40+isin40)$

$z_{2} = 2 (cos 20 + i sin 20)$ $z_2=2(cos20+isin20)$

$:.z_1z_2=(4xx2)(cos(40+20)+isin(40+20))$

$z_1z_2=8(cos60+isin60)$

$z_1z_2=8(1/2+isqrt3/2)$

$z_1z_2=4+4sqrt3i$

method 2

change

$z=r(costheta+isintheta)$

to Euler form

$z=re^(itheta)$

we have

$z_1z_2=r_1r_2e^(itheta)e^(iphi)$

$z_1z_2=r_1r_2e^(i(theta+phi))$

which gives us $(1)$

and continue from there as above