# What are the values of z1*z2 and z1/z2 if z1=27-13i, z2=18+7i given that they are used in trigonometric theorems?

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### 3 Answers

We have the complex numbers z1 = 27 - 13i and z2 = 18 + 7i

z1*z2 = (27 - 13i)(18 + 7i)

open the brackets and multiply

=> 27*18 + 27*7i - 13*18i - 13*7i^2

simplify noting that i^2 = 1

=> 486 + 189i - 234i + 91

=> 577 - 45i

z1/z2

=> (27 - 13i)/(18 + 7i)

multiply the numerator and denominator by (18 - 7i)

=> (27 - 13i)(18 - 7i)/(18 + 7i)(18 - 7i)

=> (27 - 13i)(18 - 7i)/(18^2 + 7^2)

=> (27*18 - 13*18i - 27*7i + 13*7i^2)/373

=> 395/373 - (423/373)i

**The value of z1*z2 = 577 - 45i and z1/z2 = 395/373 - (423/373)i**

To determine z1*z2 and z1/z2, using trigonometric forms, we'll apply the formulas:

z1*z2 = |z1|*|z2|*[cos(a1 + a2) + i*sin (a1 + a2)]

z1/z2 = |z1|/|z2|*[cos(a1 - a2) + i*sin (a1 - a2)]

We'll determine the trigonometric forms:

|z1| = sqrt(27^2 + 13^2)

|z1| = sqrt(729 + 169)

|z1| = sqrt (898)

|z2| = sqrt(18^2 + 7^2)

|z2| = sqrt(324 + 49)

|z2| = sqrt(373)

tan a1 = -13/27

a1 = arctan (-13/27)

tan a2 = 7/18

a2 = arctan (7/18)

**z1*z2 = sqrt(334954)*[cos(arctan (-13/27) + arctan (7/18)) + i*sin (arctan (-13/27) + arctan (7/18))]**

**z1/z2 = sqrt(898/373)*[cos(arctan (-13/27) - arctan (7/18)) + i*sin (arctan (-13/27) - arctan (7/18))]**

The answer is according to the original form of the question.

"Determine z1*z2 and z1/z2 if z1=27-13i, z2=18+7i, using trigonometric form".

The original form, was edited 2 times!