### Author Topic: Question 3c for 2020S Night Sitting  (Read 362 times)

#### Yuyan Liu

• Newbie
• Posts: 3
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##### Question 3c for 2020S Night Sitting
« on: October 27, 2020, 06:19:33 PM »
Can someone explain how to get to the answer of $cosh(z)$?

#### RunboZhang

• Sr. Member
• Posts: 51
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##### Re: Question 3c for 2020S Night Sitting
« Reply #1 on: October 27, 2020, 07:22:04 PM »
$\text{Since we have: } \ cosh(x) = cos(ix) \ \text{ and } \ sinh(x) = -i \cdot sin(ix) \\ \text{By substituting and rearranging, we have the following: }$

\begin{gather} \begin{aligned} f(x,y) &= cosh(x) \cdot sin(y) + ( - sinh(x) \cdot cos(y)) \cdot i + C \cdot i \\\\ &= cos(ix) \cdot sin(y) - sin(ix) \cdot cos(y) + C \cdot i \\\\ &= sin(y - ix) + C \cdot i \\\\ &= sin(-i \cdot (x+iy)) + C \cdot i \\\\ &= - sin(i(x+iy)) + C \cdot i \\\\ &= -i \cdot sinh(x + iy) + C \cdot i \\\\ &= -i \cdot sinh(z) + C \cdot i \\\\ \end{aligned} \end{gather}

$\text{I think the answer has a typo, I got} \ -i \cdot sinh(z) + C \cdot i \ \text{ instead of } cosh(z) + C \cdot i$
« Last Edit: October 27, 2020, 07:32:06 PM by RunboZhang »

#### Xuefeng Fan

• Jr. Member
• Posts: 11
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##### Re: Question 3c for 2020S Night Sitting
« Reply #2 on: December 07, 2020, 02:43:43 PM »
f(x,y) = coshxsiny +i(-sinhxcosy+c)
=cos(ix)siny +i(-(-isin(ix)cosy +c)
=cos(ix)siny - sin(ix)cosy +ic
=sin(y-ix)+ic
=sin(-i(x+iy))+ic
=-sin(i(x+iy)) + ic
=-isinh(x+iy) +ic
=-isinhz+ic