Calculating (a+bi)^(c+di): How to Find it?

  • Thread starter kishtik
  • Start date
In summary, the conversation is about finding the value of (a+bi)^(c+di) and how to solve for the second part, (a+bi)^di. The suggested approach is to use identities involving exponential, logarithmic, and trigonometric functions of complex numbers. The participant also mentions the De-Moivre's theorem and suggests studying these concepts to better understand and solve the problem.
  • #1
kishtik
100
0
What is (a+bi)^(c+di) ? How can I find this?
=[(a+bi)^c]x[(a+bi)^di]=? Now I can go binomial for the first part but what about (a+bi)^di?
 
Mathematics news on Phys.org
  • #2
Have you dealt with exponentials and logarithms of complex numbers yet?

The identities that'll help you are:

[tex]
\begin{array}{l}
\forall x,y \in {\rm R} \\
\exp (x + iy) = \exp (x)(\cos (y) + i\sin (y)) \\
\log _e (x + iy) = \log _e (\sqrt {x^2 + y^2 } ) + i\arctan (y/x) \\
\end{array}
[/tex]

(I've glossed over the fact that the log function is actually multi-valued ... let me know if you need this explained further).

[tex]
\begin{array}{l}
\forall z,w \in {\rm C }, w \neq 0 \\
\log _e (z^w ) = w\log _e (z) \\
z^w = \exp (w\log _e (z)) \\
\end{array}
[/tex]

See how you get on.
 
  • #3
I couldn't understand the first three quations although I did the last three. And I have no idea about expotentials and logarithms of complex numbers. Thanks.
 
  • #4
(a+ib)^{di} = ((m \exp (ni))^d)^i
m = \sqrt{a^2 + b^2}, n = arctan (b/a)
= (m \exp(ni))^i)^d
= ((m^i) \exp(-n))^d
= m^{id} \exp(-nd)
m^{id} is a complex number.



Regarding the equations,
They are pretty simple
The first equation is the famous De-Moivre's theorem. Prrof can be found in any algebra book.
For Eqn 2, from De-Moivre's theorem,
[tex]
\begin{array}{l}
x + iy = \sqrt(x^2+y^2)\exp(i\arctan(y/x))
\end{array}
[/tex]
Take logarithms on both sides and you get equation 2.
 
  • #5
I'm not sure if there are ways to solve this without exp, log etc.

Does anyone know another way?
 
  • #6
kishtik said:
I couldn't understand the first three quations although I did the last three. And I have no idea about expotentials and logarithms of complex numbers. Thanks.
Yeah, you really need to look up exponents, logs, trigonometric functions and hyperbolic functions in relationship to complex numbers otherwise you'll struggle to deal with such problems.
 

1. How do I calculate (a+bi)^(c+di)?

To calculate (a+bi)^(c+di), you can use the formula: (a+bi)^(c+di) = e^[(c+di)ln(a+bi)], where e is the base of natural logarithm.

2. What are the steps to find (a+bi)^(c+di)?

The steps to find (a+bi)^(c+di) are as follows:

Step 1: Rewrite the complex number (a+bi) in polar form as r(cosθ + isinθ).

Step 2: Apply the laws of exponent to the polar form, (r(cosθ + isinθ))^(c+di).

Step 3: Use Euler's formula to convert the exponential form to trigonometric form, r^(c+di)e^[(c+di)θ].

Step 4: Expand the expression using the properties of exponents.

Step 5: Simplify the resulting expression to get the final answer.

3. Can we use a calculator to find (a+bi)^(c+di)?

Yes, you can use a scientific calculator with complex number capabilities to find (a+bi)^(c+di). Make sure to use the correct order of operations and convert any complex numbers to polar form before inputting them into the calculator.

4. What is the significance of the imaginary unit i in (a+bi)^(c+di)?

The imaginary unit i represents the square root of -1. It is necessary to use in complex numbers as it allows us to work with numbers that are not on the real number line. In the formula (a+bi)^(c+di), i helps to create a complex number in the exponent, which results in a complex number as the final answer.

5. Can (a+bi)^(c+di) have multiple solutions?

Yes, (a+bi)^(c+di) can have multiple solutions. This is because complex numbers have an infinite number of roots. Depending on the values of a, b, c, and d, the resulting complex number can have different solutions. It is important to carefully follow the steps and simplify the expression to get the principal value of (a+bi)^(c+di).

Similar threads

Replies
3
Views
593
  • General Math
Replies
2
Views
1K
Replies
5
Views
1K
  • General Math
Replies
8
Views
1K
  • General Math
Replies
3
Views
7K
  • Precalculus Mathematics Homework Help
2
Replies
39
Views
4K
Replies
8
Views
663
Replies
1
Views
2K
Replies
3
Views
661
  • Introductory Physics Homework Help
Replies
4
Views
625
Back
Top