Given the complex number z1=12(cos⁡13π8+isin⁡13π8) and z2=2(cos⁡11π8+isin⁡11π8), express the result of z1z2 in rectangular form with fully simplified fractions and radicals.

Answered: "Given the complex number z1=12(cos⁡13π8+isin⁡13π8) and z2=2(cos⁡11π8+isin⁡11π8), express..."

chillywilly12a

2021-09-18

Given the complex number

z_{1} = 12 (\cos \frac{13 π}{8} + i \sin \frac{13 π}{8})

and

z_{2} = 2 (\cos \frac{11 π}{8} + i \sin \frac{11 π}{8})

, express the result of

\frac{z_{1}}{z_{2}}

in rectangular form with fully simplified fractions and radicals.

grbavit

Skilled2021-09-19Added 109 answers

Given two complex numbers,

z_{1} = 12 (\cos (\frac{13 π}{8}) + i \sin (\frac{13 π}{8}))

and

z_{2} = 2 (\cos (\frac{11 π}{8}) + i \sin (\frac{11 π}{8}))

We have to express

\frac{z_{1}}{z_{2}}

in rectangular form. Given complex numbers are of the form

z = r (\cos (θ) + i \sin (θ))

Changing

z = r (\cos (θ) + i \sin (θ))

to Euler form

z = r e^{t θ}

we get,

z_{1} = 12 e^{\frac{13 x}{8}}

and

z_{2} = 2 e^{\frac{11 x}{8}}

So,

z = \frac{z_{1}}{z_{2}} = \frac{12 e^{\frac{13 x}{8}}}{2 e^{\frac{11 x}{8}}}

= 6 e^{\frac{13 x}{8} - \frac{11 x}{8}}

= 6 e^{\frac{π}{4}}

= 6 (\cos (\frac{π}{4}) + i \sin (\frac{π}{4}))

= 6 (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})

= 3 \sqrt{2} (1 + i)

z = 3 \sqrt{2} (1 + i)

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