# MULTIPLYING COMPLEX NUMBERS IN EXPONENTIAL FORM

Example 1 :

Find the product of the following complex numbers :

(cos2θ + isin2θ)(cos3θ + isin3θ)

Solution :

Convert the above complex numbers to exponential form.

cos2θ + isin2θ = ei2θ

cos3θ + isin3θ = ei3θ

(cos2θ + isin2θ)(cos3θ + isin3θ) :

= ei2θ ⋅ ei3θ

= ei2θ + i3θ

= ei(2θ + 3θ)

= ei5θ

= cos5θ + isin5θ

Example 2 :

Find the product of the following complex numbers :

(cos4θ + isin4θ)-6(cosθ + isinθ)8

Solution :

Convert the above complex numbers to exponential form.

cos4θ + isin4θ = ei4θ

cosθ + isinθ = eiθ

(cos4θ + isin4θ)-6(cosθ + isinθ):

= (ei4θ)-6 ⋅ (eiθ)8

= e-i24θ ⋅ ei8θ

= e-i24θ + i8θ

= ei(-24θ + 8θ)

ei(-16θ)

= e-i16θ

= cos16θ - isin16θ

Example 3 :

Find the product of the following complex numbers :

(cos4θ + isin4θ)12(cos5θ - isin5θ)-6

Solution :

Convert the above complex numbers to exponential form.

cos4θ + isin4θ = ei4θ

cos5θ - isin5θ = e-i5θ

(cos4θ + isin4θ)12(cos5θ - i sin5θ)-6 :

= (ei4θ)12 ⋅ (e-i5θ)-6

= ei48θ ⋅ ei30θ

= ei48θ + i30θ

= ei(48θ + 30θ)

= ei78θ

= cos78θ + isin78θ

Example 4 :

Find the division of the following complex numbers

(cosα + isinα)3/(sinβ + icosβ)4

Solution :

In the above division, complex number in the denominator is not in polar form.

First, convert the complex number in denominator to polar form.

sinβ + icosβ = cos(90 - β) + isin(90 - β)

Then,

(cosα + isinα)3/(sinβ + icosβ):

= (cosα + isinα)3/[cos(90 - β) + isin(90 - β)]4

Convert the above complex numbers to exponential form.

cosα + isinα = ei4α

cos(90° - β) + isin(90° - β) = ei(90° - β)

Then,

= (ei4α)3/(ei(90° - β))4

= (ei12α)/(ei(360° - 4β))

Write the division as multiplication by changing the sign of the exponent in denominator.

= (ei12α⋅ (e-i(360° - 4β))

ei12α - i(360° - 4β)

ei[4α - (360° - 4β)]

ei[4α - 360° + 4β)]

ei[4α + 4β - 360°]

= cos(4α + 4β - 360°) + isin(4α + 4β - 360°)

= cos[-(360-(4α + 4β))] + isin[-(360-(4α + 4β))]

= cos(4α + 4β) - isin[360-(4α + 4β)]

= cos(4α + 4β) + i sin(4α + 4β)

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