(2-3i)(3+5i)
Please do not post any more questions until you've shown that you're making an attempt to solve these problems.
As it is, it looks like you're just dumping your homework here.
I have a lot more. I'm only putting one problem from each chapter so I can see the steps again.Can you help me with this then?
If I was here to just have everything answered i'd use y ahooanswer.
(2-3i)(3+5i)
These are done by the FOIL method. Remember i = -1.
http://www.purplemath.com/modules/polymult2.htm
Thanx for no steps...
That link tells you how to solve that type problem.
To multiply complex numbers, you can use the distributive property.
Let's multiply (2 - 3i) and (3 + 5i) step by step:
Step 1: Multiply the real components.
(2 - 3i) * 3 = 6 - 9i
Step 2: Multiply the imaginary components.
(2 - 3i) * 5i = -15i + 10i^2
Step 3: Simplify the result.
i^2 is defined as -1, so i^2 = -1.
Therefore, -15i + 10i^2 = -15i - 10 = -10 - 15i
Step 4: Combine the real and imaginary components.
(6 - 9i) + (-10 - 15i) = (6 - 10) + (-9i - 15i)
= -4 - 24i
So, (2 - 3i)(3 + 5i) = -4 - 24i.