With steps please?

Question

890 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-01T20:48:49

Frankly, I don't know what the degree symbols are for.
Here's what I would do:

#z_1*z_2 = (e^(5^oi)) * (2*e^(2^oi)) = #

(just rearranging terms) #(e^(5^oi)) * (2)*(e^(2^oi)) = #

#(2)(e^(5^oi))(e^(2^oi))#.

When you multiply two numbers raised to an exponent and the bases are the same, you add the exponents, for example:

#x^3 * x^8 = x^11#

Here, #(2)*(e^(5^o*i)) *(e^(2^o*i)) = #

#(2)* e^(((5^oi) + (2^oi)))#.

#(5^oi)# and #(2^oi)# are like terms, so adding them equals:

#7^oi#.

This gives us:

#z_1z_2 = 2e^(7^oi)#.

If this is not correct or is not what is intended with the degree symbols, please let me know.