# How do you evaluate #1200=300(1+r)^{5}#?

##### 2 Answers

#### Explanation:

#1200=300(1+r)^5#

First, divide both sides of the equation by

#1200/300=(300(1+r)^5)/300#

#4=(1+r)^5#

To undo the power of

#4^(1/5)=((1+r)^5)^(1/5)#

#4^(1/5)=1+r#

Finally, subtract

#r=4^(1/5)-1#

Personally, I prefer the simplification

#r=2^(2/5)-1#

#### Explanation:

We have:

# 1200=300(1+r)^5 #

Let

# 1200 = 300z^5 => z^5 = 4 #

First, we will put the equation into polar form:

# |omega| = 4#

# theta =0#

So then in polar form we have:

# z^5 = 4(cos0 + isin0) #

We now want to solve the equation

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of

# z^5 = 4(cos(0+2npi) + isin(0+2npi)) \ \ \ n in ZZ #

By De Moivre's Theorem we can write this as:

# z = (4(cos(0+2npi) + isin(0+2npi)))^(1/5) #

# \ \ = 4^(1/5)(cos(2npi) + isin(2npi)))^(1/5) #

# \ \ = 4^(1/5)(cos((2npi)/5) + isin((2npi)/5)) #

# \ \ = 4^(1/5)(costheta + isintheta) #

Where:

# theta =(2npi)/5#

And we will get

After which the pattern continues (due the above mentioned periodicity).

We can plot these solutions on the Argand Diagram: