If x is real, show that (2 + j)e^((1 + j3)x) + (2 − j)e^((1 − j3)x) is also real


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Question Number 51284 by Tawa1 last updated on 25/Dec/18

$If x is real, show that (2 + j) e^{(1 + j 3) x} + (2 - j) e^{(1 - j 3) x}$ $is also real$
Commented by maxmathsup by imad last updated on 25/Dec/18

$f i r s t w h a t m e a n j a n d j 3 ?$
	Answered by mr W last updated on 25/Dec/18
	$2+i=(√5)((2/(√5))+(1/(√5))i)=(√5)(cos α+i sin α) =(√5)e^(αi) with α=tan^(−1) (1/2) 2−i=(√5)((2/(√5))−(1/(√5))i)=(√5)[cos (−α)+i sin (−α)] =(√5)e^(−αi) S=(2+i)e^((1+3i)x) +(2−i)e^((1−3i)x) =(√5)e^(αi) e^((1+3i)x) +(√5)e^(−αi) e^((1−3i)x) =(√5)e^x [e^((3x+α)i) +e^(−(3x+α)i) ] =(√5)e^x [cos (3x+α)+i sin (3x+α)+cos {−(3x+α)}+i sin {−(3x+α)}] =(√5)e^x [cos (3x+α)+i sin (3x+α)+cos (3x+α)−i sin (3x+α)] =(√5)e^x [2cos (3x+α)] =2(√5)e^x cos (3x+α) =2(√5)e^x (cos α cos 3x−sin α sin 3x) =2(√5)e^x ((2/(√5))cos 3x−(1/(√5))sin 3x) =2e^x (2 cos 3x−sin 3x) =real$
	$2 + i = \sqrt{5} (\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}} i) = \sqrt{5} (\cos α + i \sin α)$ $= \sqrt{5} e^{α i}$ $w i t h α = \tan^{- 1} \frac{1}{2}$ $2 - i = \sqrt{5} (\frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} i) = \sqrt{5} [\cos (- α) + i \sin (- α)]$ $= \sqrt{5} e^{- α i}$ $S = (2 + i) e^{(1 + 3 i) x} + (2 - i) e^{(1 - 3 i) x}$ $= \sqrt{5} e^{α i} e^{(1 + 3 i) x} + \sqrt{5} e^{- α i} e^{(1 - 3 i) x}$ $= \sqrt{5} e^{x} [e^{(3 x + α) i} + e^{- (3 x + α) i}]$ $= \sqrt{5} e^{x} [\cos (3 x + α) + i \sin (3 x + α) + \cos {- (3 x + α)} + i \sin {- (3 x + α)}]$ $= \sqrt{5} e^{x} [\cos (3 x + α) + i \sin (3 x + α) + \cos (3 x + α) - i \sin (3 x + α)]$ $= \sqrt{5} e^{x} [2 \cos (3 x + α)]$ $= 2 \sqrt{5} e^{x} \cos (3 x + α)$ $= 2 \sqrt{5} e^{x} (\cos α \cos 3 x - \sin α \sin 3 x)$ $= 2 \sqrt{5} e^{x} (\frac{2}{\sqrt{5}} \cos 3 x - \frac{1}{\sqrt{5}} \sin 3 x)$ $= 2 e^{x} (2 \cos 3 x - \sin 3 x)$ $= r e a l$
	Commented by Tawa1 last updated on 25/Dec/18

	$God bless you sir$
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