What is the trigonometric form of $(4 - 4 i)$ $(4-4i)$ ?

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Answer 1 · 2016-03-15T10:21:47

$= 4 \sqrt{2} [cos (- \frac{3 π}{4}) + i sin (- \frac{3 π}{4})]$ $=4sqrt2[cos(-(3pi)/4)+isin(-(3pi)/4)]$

The trigonometric form looks like this : $r (cos θ + i sin θ)$ $r(costheta+isintheta)$

where $r$ $r$ is the modulus and $θ$ $theta$ the arguement.

Step 1:
find the modulus(magnitude)
$r = \sqrt{{(4)}^{2} + {(- 4)}^{2}} = 4 \sqrt{2}$ $r=sqrt((4)^2+(-4)^2)=4sqrt(2)$

Step 2:
find the arguement.
$θ = arctan (\frac{4}{4}) - π = \frac{π}{4} - π = - \frac{3 π}{4}$ $theta= arctan(4/4)- pi= pi/4-pi=-(3pi)/4$

Step 3:
$z = 4 \sqrt{2} [cos (- \frac{3 π}{4}) + i sin (- \frac{3 π}{4})]$ $z=4sqrt2[cos(-(3pi)/4)+isin(-(3pi)/4)]$

What is the trigonometric form of (4−4i) (4-4i) ?