Find all solutions of sinz = 100 . Find all solutions of sinhz = - i (Cheek page

Question

Find all solutions of sinz = 100 . Find all solutions of sinhz = - i (Cheek pages 21 and 22 of lecture 5 and "Solving complex trigonometric equation" under non-narrated Lecture 5 and examples) Convert time varying functions i1 = 6cos(377t-pi/3) and i2 = -9sin(377t+pi/4) into complex phasors I1 and I2 We know i1+i2=Acos(377t+phi). Find A and phi (in radian) with the phasor method. Complex power is defined as P =|I1 + I2| (R + iomegaL) where I1 and I2 are phasors in a) and R=100 and L=1. Compute P in polar form. (Check page 28 in lecture 5 and "Example sum and product of complex number and phasor Extra-Credit: Voltage v(t)= 10sin(200t + 60degree) (v) and current i(t) = 20cos(200t - 45degree)(A) Find phasor form of voltage (V) and current (I). Compute complex power P = VI and average power Pavg = Re {p} / 2 Find the time varying cos function for the following phasors that have angular frequency of omega: 6

Explanation / Answer

2) i) sin(z) = sin(x+iy) = sin(x)cos(iy) + cos(x)sin(iy) = 100

Now use the identities cos(ix)=cosh(x) and sin(iy)=isinh(y). ( these can be derived from the definitions of sin and cos in terms of exp ).

Equation becomes sin(x)cosh(y) + icos(x)sinh(y) = 100 … (i)

Equating imaginary parts : cos(x)sinh(y)=0 so cos(x)=0 or sinh(y)=0

If sinh(y)=0 then y=0 and (i) becomes sin(x)=100 which is impossible. (x is real)

Hence cos(x)=0 which implies that x=n+/2 for integer n.

Subbing in (i) gives (1)cosh(y)=100

Since cosh(y)>0 always, n must be even and in this case cosh(y)=100

To solve cosh(y)=100 use calculator to find cosh¹(100)=±5.2983. ( inspect graph of cosh to see there are two solutions )

Bringing all this together gives complex solutions of sin(z)=100 as

z = (4m+1)/2 ± 5.2983i for m any integer.

2)ii) sinh(z) = sinh(x+iy) = sinh(x)cosh(iy)+cosh(x)sinh(iy)

= -sinh(x)*cos(y) - icosh(x)*sin(y) = -i

so sinh(x) *cos(y) = 0

cos(y) = 0

y = (2n+1)pi/2

and cosh(x)*sin(y) = 1

so cosh(x) = +-1

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