How do you write the following in trigonometric form and perform the operation given $(2 + 2 i) (1 - i)$ $(2+2i)(1-i)$ ?

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How do you write the following in trigonometric form and perform the operation given $(2 + 2 i) (1 - i)$ $(2+2i)(1-i)$ ?

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Answer 1 · 2016-10-20T12:09:25

$(2 + 2 i) (1 - i) = 4$ $(2+2i)(1-i)=4$

Explanation:

Let $z_{1} = 2 + 2 i$ $z_1=2+2i$ And $z_{2} = 1 - i$ $z_2=1-i$
Then the moduli are $∣ ∣ z_{1} ∣ = \sqrt{2^{2} + 2^{2}} = 2 \sqrt{2}$ $∣z_1∣ = sqrt (2^2+2^2)=2sqrt2$
and $z_{2} = \sqrt{1 + 1} = \sqrt{2}$ $z_2= sqrt(1+1)=sqrt2$
Rewrite the numbers
$z_{1} = 2 + 2 i = 2 \sqrt{2} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = 2 \sqrt{2} ({cos θ}_{1} + i {sin θ}_{1})$ $z_1=2+2i=2sqrt2(1/sqrt2+i/sqrt2)=2sqrt2(costheta_1+isintheta_1)$
Here $θ_{1} = \frac{π}{4}$ $theta_1=pi/4$
so $z_{1} = 2 \sqrt{2} (cos (\frac{π}{4}) + i sin (\frac{π}{4})) = 2 \sqrt{2} e^{i \frac{π}{4}}$ $z_1=2sqrt2(cos(pi/4)+isin(pi/4))= 2sqrt2e^(ipi/4)$
Similarly
$z_{2} = 1 - i = \sqrt{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \sqrt{2} ({cos θ}_{2} - i {sin θ}_{2})$ $z_2=1-i=sqrt2(1/sqrt2-i/sqrt2)=sqrt2(costheta_2-isintheta_2)$
$θ_{2} = - \frac{π}{4}$ $theta_2=-pi/4$

so $z_{2} = \sqrt{2} (cos (- \frac{π}{4}) + i sin (- \frac{π}{4})) = \sqrt{2} e^{- i \frac{π}{4}}$ $z_2=sqrt2(cos(-pi/4)+isin(-pi/4))= sqrt2e^(-ipi/4)$
so the result is $z_{1} \cdot z_{2} = 2 \sqrt{2} e^{i \frac{π}{4}} \cdot \sqrt{2} e^{- i \frac{π}{4}} = 4$ $z_1*z_2=2sqrt2e^(ipi/4)*sqrt2e^(-ipi/4)=4$

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How do you write the following in trigonometric form and perform the operation given $(2 + 2 i) (1 - i)$ $(2+2i)(1-i)$ ?

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How do you write the following in trigonometric form and perform the operation given (2+2i)(1−i)(2+2i)(1-i)?

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How do you write the following in trigonometric form and perform the operation given $(2 + 2 i) (1 - i)$ $(2+2i)(1-i)$ ?