Mathcalc8 Basis : complex number calculation

In this part two of the post, we will concentrate on enter formulas that contain complex numbers. Complex number is the basic element of this calculator. It is also the reason why we call our calculator math calculator, Mathcalc8. Nearly all functions support complex number. In this part two of the post, we will include not just the input hints but also the answers, both from Mathcalc8 and textbook for the calculations. If you are a K12 college student, you may find some of the calculations very tricky and 'out of syllabus'. So, for those calculations, a brief note will be given as a background reference. It is also noticed that some of the calculations need the full version of Mathcalc8. For those users with free version, you can see them as demo. Now, Let's begin :

1. z1 = -5+6i , z2 = 1-2i, what is z1+z2 ?
   Input : -5+6*sqrt(-1)+1-2*sqrt(-1)
   Or
   Input : complex(-5,6)+complex(1,-2)
   Ans. -4+4i
    
2. what is z1 x z2 ?
   Input : (-5+6*sqrt(-1))*(1-2*sqrt(-1))
   Or
   Input : complex(-5,6)*complex(1,-2)
   Ans. 7+16i
    
3. 5+5i     20
   —— + ——
   3-4i     4+3i
   Input : (5+5*sqrt(-1))/(3-4*sqrt(-1))+20/(4+3*sqrt(-1))
   Or
   Input : complex(5,5)/complex(3,-4)+20/complex(4,3)
   Ans. 3-i
    
4. (1+i)¹⁰⁰⁰
   Input : pow(1+sqrt(-1),1000)
   Ans. 3.2734e150 (Mathcalc8)
   
   Ans. 2⁵⁰⁰ (textbook)
    
5. ((1+i)÷(1-i))¹⁶+((1-i)÷(1+i))⁸
   Input : pow(complex(1,1)/complex(1,-1),16)+pow(complex(1,-1)/complex(1,1),8)
   Ans. 2
    
6. z = 4+3i , what is its magnitude and angle appears in complex plane ?
   Input : (mag(4+3*sqrt(-1)),angle(4+3*sqrt(-1)))
   Ans. 5,36.87^o
   Comment : a list is written in input instead of a single formula. In later post, we will further discuss this technique.
    
7. 4^½
   Input 1 : sqrt(4)
   Or
   Input 2 : pow(4,1/2)
   Or
   Input 3 : root(4,1/2)
   Ans. 2 (input 1)
   Ans. 2 (input 2)
   Ans. 2,-2 (input 3)
   Comment : the function of sqrt is actually the short handed form of pow( ,1/2), so both sqrt and pow give same answer of 2. The function pow is more powerful than sqrt because its index is not limited to 1/2. However, the most powerful function is root which gives all the possible solution including -2.
    
8. 1^1/3
   Input : root(1,1/3)
   Ans. 1,-0.5+0.866i,-0.5-0.866i
   Comment : if pow is used, the answer will be 1.
    
9. (-7+24i)^1/4
   Input : root(-7+24*sqrt(-1),1/4)
   Ans. 2+i,-1+2i,1-2i,-2-i
   Comment : if pow is used, only one solution -2-i resulted.
    
10. Ln(i)
    Input : ln(sqrt(-1))
    Ans. 1.5708i (Mathcalc8)
    Comment : The function Ln is the function of ln which is expanded to cover complex number. More detail will be given on the note below. The value given by Mathcalc8 is the principal value of the formula.
     
11. (2 ÷ i) x Ln(((5+i)⁴ x (-239+i)) ÷ ((5-i)⁴ x (-239-i)))
    Input : 2/sqrt(-1)*ln(pow(5+sqrt(-1),4)*(-239+sqrt(-1))/(pow(5-sqrt(-1),4)*(-239-sqrt(-1))))
    Ans. 3.1415927 (Mathcalc8)
    Comment : The answer is actually π.
     
12. iⁱ
    Input : pow(sqrt(-1),sqrt(-1))
    Ans. 0.20787 (Mathcalc8)
    Comment : The answer is also the principal value of the formula.
     
13. cos(i)
    Input : cos(i)
    Ans. 1.54308 (Mathcalc8)
    Comment : More detail is given on the below note.
     

Note :

Definition of Ln(z) :

z = e^w where w is defined as Ln(z)
z = |z|e^iθ = e^ln(|z|)e^iArg(z)
Ln(z) = w = ln(|z|) + iArg(z) + 2kπ where k = 0, ±1, ±2 ...
The value of Ln(z) at k = 0 is called the principal value (pv) of Ln(z).
pv of Ln(i) = ln(|i|) + iArg(i) = 0 + i π/2 = i π/2
iⁱ = (e^Ln(i))ⁱ = e^{(iπ/2+2kπ) x i} = e^-π/2+2kπi
pv of iⁱ = e^-π/2

Definition of cos(z), sin(z) :

e^iz = cos(z) + i sin(z) ( Extension from Euler's formula e^iθ = cos(θ) + i sin(θ) )
e^-iz = cos(z) - i sin(z)
cos(z) = (e^iz+e^-iz)/2
sin(z) = (e^iz-e^-iz)/(2i)

cos(i) = (e^ixi+e^-ixi)/2 = (e^-1+e¹)/2