Multiplication of Complex Numbers
Key Questions
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Let
#z_1# and#z_2# be two complex numbers.By rewriting in exponential form,
#{(z_1=r_1e^{i theta_1}),(z_2=r_2 e^{i theta_2}):}# So,
#z_1 cdot z_2 =r_1e^{i theta_1}cdot r_2 e^{i theta_2} =(r_1 cdot r_2)e^{i(theta_1+theta_2)}# Hence, the product of two complex numbers can be geometrically interpreted as the combination of the product of their absolute values (
#r_1 cdot r_2# ) and the sum of their angles (#theta_1+theta_2# ) as shown below.
I hope that this was clear.
Wataru
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1
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Dec 12 2014
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In trigonometric form, a complex number looks like this:
#a + bi = c*cis(theta)#
where#a# ,#b# and#c# are scalars.Let two complex numbers:
#-> k_(1) = c_(1)*cis(alpha)#
#-> k_(2) = c_(2)*cis(beta)#
#k_(1)*k_(2) = c_(1) * c_(2) * cis(alpha) * cis(beta) =#
#= c_(1) * c_(2) * (cos(alpha) + i*sin(alpha)) * (cos(beta) + i*sin(beta))# This product will end up leading to the expression
#k_(1)*k_(2) =#
#= c_(1)*c_(2)*(cos(alpha + beta) + i*sin(alpha + beta)) =#
#= c_(1)*c_(2)*cis(alpha+beta)# By analyzing the steps above, we can infer that, for having used generic terms
#c_(1)# ,#c_(2)# ,#alpha# and#beta# , the formula of the product of two complex numbers in trigonometric form is:
#(c_(1) * cis(alpha)) * (c_(2) * cis(beta)) = c_(1)*c_(2)*cis(alpha+beta)# Hope it helps.
Bernardo Russo G. Ferreira · 1 · Oct 2 2014
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To explain this, I will name two generic complex.
#c_1 = a*cis(alpha)# and#c_2 = b*cis(beta)# The product between
#c_1# and#c_2# is:
#ab*cis(alpha)cis(beta) =#
#ab*(cos(alpha)+isin(alpha)) (cos(beta)+isin(beta)) =#
#ab*({cos(alpha)cos(beta)-sin(alpha)sin(beta)} +#
#{i(sin(alpha)sin(beta)+cos(alpha)sin(beta)}) =#
#ab*{cos(a+b)+isin(a+b)}# //Therefore, we can assume that the product of the two complex numbers
#c_1# and#c_2# can be generaly given by the form above.Ex.:
#(2*cis(pi)) * (3*cis(2pi)) = 6*cis(3pi) = 6*cis(pi)# Hope it helps.
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Answer:
Hope this helps.
Explanation:
Step 1: Distribute (or FOIL) to remove the parenthesis.
Step 2 : Simplify the powers of i, specifically remember that#i^2 = –1# .
Step 3 : Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers.Example 1 – Multiply: (4 – 3i)(2 + 5i)
Step 1:
#"Distribute to remove the parenthesis. " color(green)(8 + 20 i - 6 i - 15 i^2# Step 2:
#"Simplify the powers of i, specifically remember that " i^2 = –1#
#color(blue)(8 + 20 i - 6 i + 15.Step 3:
#"Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers."#
#color(crimson)(23 + 14 i# Example 2 – Multiply: (7 – 9i)(4 - 6i)
Step 1:
#"Distribute to remove the parenthesis. " color(green)(28 - 42 i - 36 i + 54 i^2# Step 2:
#"Simplify the powers of i, specifically remember that " i^2 = –1#
#color(blue)(28 - 42 i - 36 i - 54.Step 3:
#"Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers."#
#color(crimson)(-26 - 78 i# Example 3 – Multiply: (7 - 9i)(4 - 6i)(2 + 5i)
#"Step 1: Distribute (or FOIL) using only the first two complex numbers " color(green)((28 - 42 i - 36 i + 54 i^2) * (2 + 5 i)# #"Step 2: Simplify the powers of i, specifically remember that i2 = –1" color(blue)(28 - 42 i - 36 i - 54)(2 + 5i)# #"Step 3: Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers."#
#color(crimson)((-26 - 78 i) * (2 + 5 i)# #"Step 4 : Distribute to remove the parenthesis. " color(chocolate)(-52 - 130 i - 156 i - 390 i^2# #"Step 5: Simplify the powers of i, specifically remember that " i^2 = –1#
#color(magenta)(-52 - 130 i - 156 i + 390# .#"Step 6: Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers."#
#color(purple)(328 - 286 i# 
sankarankalyanam
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1
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Jul 21 2018
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First, complex numbers can come in a variety of forms!
Ex: multiply
#3i*-4i =# Remember, with multiplication you can rearrange the order (called the Commutative Property):
#3*-4*i*i =-12i^2# ... and then always substitute -1 for
#i^2# :#-12*-1 = 12# Ex: the numbers might come in a radical form:
#sqrt(-3)*4sqrt(-12) =# You should always "factor" out the imaginary part from the square roots like this:
#sqrt(-1)sqrt(3)*4*sqrt(-1)sqrt(4)sqrt(3) =# and simplify again:
#=i*4*sqrt(3)*sqrt(3)*sqrt(4)#
#=i*4*3*2 = 24i# Ex: what about the Distributive Property?
#3i(4i - 6) =# #=12i^2- 18i#
#=12(-1) - 18i#
#= -12 - 18i# And last but not least, a pair of binomials in a + bi form:
Ex: (3 - 2i)(4 + i) =
=12 + 3i - 8i -
#2i^2#
= 12 - 2(-1) + 3i - 8i
= 12 + 2 - 5i
= 14 - 5i
MrsRudolph221
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Sep 4 2014