Part A
The given expression is:
We expand to get:
Note that 
Part Bi)
The given expression is;
We simplify to get:
Note that:
This is now of the form 
, where 
.
This explains why it is a complex number;
Part bii)
The given expression is :
We simplify to get
Part C
We want to simplify:
We rewrite in terms of 
21−=2(2−)=2cos(−1)+2 sin(−1)
−1+2=−1(2)=−1(cos2+sin2)=cos2+ sin2
Is the above the correct way to write 21− and −1+2 in the form +? I wasn't sure if I could change Euler's formula to =cos()+sin(), where is a constant.
complex-numbers
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edited Mar 6 '17 at 4:38
Richard Ambler
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asked Mar 6 '17 at 3:34
14wml
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1 Answer
1
No. It is not true that =cos()+sin(). Notice that
1=1≠cos()+sin(),
for example consider this at =0.
As a hint for figuring this out, notice that
+=ln(+)
then recall your rules for logarithms to get this to the form (+)ln().
Answer:
12 is the correct answer
Step-by-step explanation:
1) so whenever we're dividing with fractions we need to change the division sign to multiplication and turn the second fraction upside down (reciprocal)
so
6*2/1=12
Julio did change the division sign to multiplication but he forgot to take the reciprocal of the second fraction (he left the second fraction as is)
what Julio did: 6*1/2=3
Hope this helps!
