The subtraction of complex numbers 
is cos(π)+i sin(π).
Given 
[cos(3π/4+i sin(3π/4) and 
=cos (π/2) +i sin(π/2)
We have to find the value of .
A complex number is a number that includes real number as well as a imaginary unit in which 
. It looks like a+ bi.
We have to first solve 
and then we will be able to find the difference.
[ cos (3π/4)+i sin (3π/4)]
[cos(π-π/4)+ i sin (π-π/4)]
= [-cos(π/4)+sin (π/4)]
=(-1/+1/)
=
=0
cos(π/2)+i sin (π/2)
=0+i*1
=1
Now putting the values of 
,
=-1
=-1+i*0
=cos (π)+i sin(π)
Hence the value of difference between is cos(π)+i sin(π).
Learn more about complex numbers at brainly.com/question/10662770
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I got 76 for the missing value
Answer:
Step-by-step explanation:
In order to make d the subject of the formula, you need to isolate it.
- You started with d-7 = 4d + 3/e
- Move 4d to the left side by subtracting 4d from both sides to cancel it from the right so you have...
d - 7 = 4d + 3/e This will leave you left with -3d - 7 = 3/e
-4d -4d
- Then move over the -7 by adding 7 to both sides...
-3d - 7 = 3/e This will leave you left with -3d = 3/e + 7
+7 +7
- Finally to get d by itself divide both sides of the equation by -3 and you'll be left with...
- You can cancel out the 3 in the -3/3e and make it -1/e so your final answer will be
The value of b is -6.
Explanation:
The expression is 
To determine the value of b, we shall solve the expression.
Applying exponent rule, 
, we get,
Applying exponent rule, 
, we have,
The expression is of the form, 
then 
Applying this rule, we get,
Dividing both sides by 4, we have,
Hence, the value of b is -6.
