How do you simplify $4 (cos (\frac{9 π}{4}) + i sin g (\frac{9 π}{4})) \div 2 [cos (- \frac{π}{2}) + i sin (- \frac{π}{2})]$ $4(cos((9pi)/4)+ising((9pi)/4))div2[cos(-pi/2)+isin(-pi/2)]$ and express the result in rectangular form?

Question

How do you simplify $4 (cos (\frac{9 π}{4}) + i sin (\frac{9 π}{4})) \div 2 [cos (- \frac{π}{2}) + i sin (- \frac{π}{2})]$ $4(cos((9pi)/4)+isin((9pi)/4))div2[cos(-pi/2)+isin(-pi/2)]$ and express the result in rectangular form?

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Answer 1 · 2017-03-24T13:02:21

$- \sqrt{2} (1 - i) .$ $-sqrt2(1-i).$

Recall that, $r (cos θ + i sin θ)$ $r(costheta+isintheta)$ is also denoted, as $r c i s θ,$ $rcistheta,$ &,

$r_{1} c i s α \div r_{2} c i s β = (\frac{r_{1}}{r_{2}}) c i s (α - β)$ $r_1cisalpha-:r_2cisbeta=(r_1/r_2)cis(alpha-beta)$ .

Using these, we find, that,

$The given Exp.= 4 c i s (9 \frac{π}{4}) \div 2 c i s (- \frac{π}{2})$ $"The given Exp.="4cis(9pi/4)div2cis(-pi/2)$

$= (\frac{4}{2}) [c i s {(9 \frac{π}{4}) - (- \frac{π}{2})}] .$ $=(4/2)[cis{(9pi/4)-(-pi/2)}].$

$= 2 {c i s (9 \frac{π}{4} + \frac{π}{2})} .$ $=2{cis(9pi/4+pi/2)}.$

$= 2 c i s (11 \frac{π}{4}) .$ $=2cis(11pi/4).$

$= 2 {cos (11 \frac{π}{4}) + i sin (11 \frac{π}{4})} .$ $=2{cos(11pi/4)+isin(11pi/4)}.$

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

$= 2 {- cos (\frac{π}{4}) + i sin (\frac{π}{4})} .$ $=2{-cos(pi/4)+isin(pi/4)}.$

$= 2 (- \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) .$ $=2(-1/sqrt2+i1/sqrt2).$

$:." The Exp.="-sqrt2(1-i).$

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