MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...
MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...
MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...
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v y = u x = cos x cosh(−y), −v x = u y = c − sin x sinh(−y) =⇒ v = − cos x sinh(−y) + h(x)<br />
=⇒ v x = sin x sinh(−y) + dh<br />
dx<br />
=⇒ h(x) = c<br />
=⇒ v = − cos x sinh(−y) + c.<br />
When c = 1, u = sin x cosh(y)<br />
v y = u x = cos x cosh(y), −v x = u y = c sin x sinh(y) =⇒ v = cos x sinh(y) + h(x)<br />
=⇒ v x = − sin x sinh(y) + dh<br />
dx<br />
=⇒ h(x) = c<br />
=⇒ v = cos x sinh(y) + c.<br />
24. u = ax 3 + by 3 harmonic =⇒ u xx + u yy = 0.<br />
u x = 3ax 2 , u y = 3by 2<br />
=⇒ u xx = 6ax and u yy = 6by =⇒ u xx + u yy = 0 and ax + by = 0 =⇒ a = b = 0.<br />
=⇒ u x = v y = 0 =⇒ v = h(x) =⇒ v x = dh<br />
dx<br />
=⇒ h(x) = c =⇒ v = c is constant.<br />
13.5<br />
1. e z = e x+iy = e x .e iy = e x (cos y + i sin y)<br />
u = e x cos y and v = e x sin y =⇒ u x = e x cos y, u y = −e x sin y, v x = e x sin y and v y = e x cos y.<br />
=⇒ u x = v y and u y = −v x .<br />
=⇒ e z is analytic for every z. =⇒ e z is entire.<br />
2-8 Values of e z :<br />
2. z = 3 + πi ⇒ e 3+πi = e 3 (cos π − i. sin π) = −e 3 ∼ = −20, 086 and |e 3+πi | ∼ = 20, 086.<br />
3. e 1+2i = e(cos 2 + i. sin 2) and |e 1+2i | = e √ cos 2 2 + i. sin 2 2.<br />
4. e √ 2− π 2 i = e √2 [cos( −π<br />
−π<br />
2<br />
) + i. sin(<br />
2 )] = −e√2 i ∼ = −4, 11325i and |e √ 2− π 2 i | ∼ = 4, 11325.<br />
5. e 7πi/2 = e 0 [cos(7πi/2) + i. sin(7πi/2)] = i and |e 7πi/2 | = |i| = 1.<br />
6. e (1+i)π = e π (cos π + i sin π) = −e π ∼ = −23, 1407 and |e (1+i)π | ∼ = 23, 1407.<br />
7. e 0,8−5i = e 0,8 .(cos(−5) + i. sin(−5)) = 2, 23.(0, 28 + i.0, 95) and |e 0,8−5i | ∼ = 2, 2.<br />
8. e 9πi/2 = e 0 [cos(9πi/2) + i. sin(9πi/2)] = i and |e 9πi/2 | = |i| = 1.<br />
9-12 Real and Imaginary parts: Let z = x + iy.<br />
9. e −2z = e −2x−2yi = e −2x (cos(−2y) + i. sin(−2y))<br />
⇒ Re(e −2z ) = e −2x (cos(2y) and Im(e −2z ) = −e −2x (sin(2y).<br />
10. e z3 = e (x+iy)3 = e (x3 −3xy 2 )+(3x 2 y−y 3 )i = e (x3 −3xy 2) (cos(3x 2 y − y 3 ) + i. sin(3x 2 y − y 3 ))<br />
⇒ Re(e z3 ) = e (x3 −3xy 2) cos(3x 2 y − y 3 ) and Im(e z3 ) = e (x3 −3xy 2) sin(3x 2 y − y 3 ).<br />
11. e z2 = e (x+iy)2 = e x2 −y 2 +2xyi = e x2 −y 2 (cos(2xy) + i. sin(2xy)).<br />
12. e 1 z = e 1<br />
x+iy<br />
= e x−iy<br />
x 2 +y 2 = e x<br />
x 2 +y 2 [cos(<br />
y<br />
) − i. sin( )].<br />
x 2 +y 2 x 2 +y 2<br />
17<br />
y