Simplify the following and leave your answer in standard form: (1-i)6 x (sqrt(3)cis(-30^))42i-1
|Mathematics
Simplify the following and leave your answer in standard form: (1-i)6 x (sqrt(3)cis(-30^))42i-1
ScanSolve AI Answer
Answer
\sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
Fresh day sthamntungwa, let's solve.
Here's the solution for Question 1.2.
Question 1.2
Simplify the following and leave your answer in standard form:
2i−1(1−i)6×(3cis(−30∘))4
Step 1: Convert (1−i) to polar form and calculate (1−i)6.
First, convert 1−i to polar form r(cosθ+isinθ).
The modulus is r=12+(−1)2=1+1=2.
The argument is θ=arctan(1−1)=−45∘ (since 1−i is in the fourth quadrant).
So, 1−i=2cis(−45∘).
Now, use De Moivre's Theorem to calculate (1−i)6:
(1−i)6=(2cis(−45∘))6=(2)6cis(6×−45∘)=23cis(−270∘)=8cis(−270∘)
Since −270∘ is equivalent to 90∘ (i.e., −270∘+360∘=90∘):
=8cis(90∘)
Convert back to rectangular form:
=8(cos90∘+isin90∘)=8(0+i(1))=8i
Step 2: Calculate (3cis(−30∘))4.
Use De Moivre's Theorem:
(3cis(−30∘))4=(3)4cis(4×−30∘)=9cis(−120∘)
Convert to rectangular form:
=9(cos(−120∘)+isin(−120∘))=9(−21−i23)=−29−i293
Step 3: Multiply the results from Step 1 and Step 2 to get the numerator.
Numerator =(8i)×(−29−i293)=8i(−29)+8i(−i293)=−36i−i22723
Since i2=−1:
=−36i−(−1)(363)=363−36i
Step 4: Simplify the denominator.
The denominator is 2i−1, which can be written as −1+2i.
Step 5: Divide the numerator by the denominator and express in standard form.
−1+2i363−36i
Multiply the numerator and denominator by the conjugate of the denominator, which is −1−2i:
(−1+2i)(−1−2i)(363−36i)(−1−2i)
Calculate the denominator:
(−1)2−(2i)2=1−4i2=1−4(−1)=1+4=5
Calculate the numerator:
(363−36i)(−1−2i)=363(−1)+363(−2i)−36i(−1)−36i(−2i)=−363−723i+36i+72i2=−363−723i+36i−72
Group the real and imaginary parts:
=(−363−72)+(36−723)i
Now, combine the numerator and denominator:
5(−363−72)+(36−723)i=5−363−72+536−723i
The final answer in standard form is ∗5−363−72+536−723i.
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ScanSolve AI Answer
Fresh day sthamntungwa, let's solve. Here's the solution for Question 1.2. Question 1.2 Simplify the following and leave your answer in standard form: (1-i)^6 × (sqrt(3)cis(-30^))^42i-1 Step 1: Convert (1-i) to polar form and calculate (1-i)^6. First, convert 1-i to polar form r( + i). The modulus is r = sqrt(1^2 + (-1)^2) = sqrt(1+1) = sqrt(2). The argument is = ((-1)/(1)) = -45^ (since 1-i is in the fourth quadrant). So, 1-i = sqrt(2)cis(-45^). Now, use De Moivre's Theorem to calculate (1-i)^6: (1-i)^6 = (sqrt(2)cis(-45^))^6 = (sqrt(2))^6 cis(6 × -45^) = 2^3 cis(-270^) = 8 cis(-270^) Since -270^ is equivalent to 90^ (i.e., -270^ + 360^ = 90^): = 8 cis(90^) Convert back to rectangular form: = 8( 90^ + i 90^) = 8(0 + i(1)) = 8i Step 2: Calculate (sqrt(3)cis(-30^))^4. Use De Moivre's Theorem: (sqrt(3)cis(-30^))^4 = (sqrt(3))^4 cis(4 × -30^) = 9 cis(-120^) Convert to rectangular form: = 9((-120^) + i(-120^)) = 9(-(1)/(2) - isqrt(3)2) = -(9)/(2) - i9sqrt(3)2 Step 3: Multiply the results from Step 1 and Step 2 to get the numerator. Numerator = (8i) × (-(9)/(2) - i9sqrt(3)2) = 8i(-(9)/(2)) + 8i(-i9sqrt(3)2) = -36i - i^2 72sqrt(3)2 Since i^2 = -1: = -36i - (-1)(36sqrt(3)) = 36sqrt(3) - 36i Step 4: Simplify the denominator. The denominator is 2i-1, which can be written as -1+2i. Step 5: Divide the numerator by the denominator and express in standard form. 36sqrt(3) - 36i-1 + 2i Multiply the numerator and denominator by the conjugate of the denominator, which is -1-2i: (36sqrt(3) - 36i)(-1 - 2i)(-1 + 2i)(-1 - 2i) Calculate the denominator: (-1)^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5 Calculate the numerator: (36sqrt(3) - 36i)(-1 - 2i) = 36sqrt(3)(-1) + 36sqrt(3)(-2i) - 36i(-1) - 36i(-2i) = -36sqrt(3) - 72sqrt(3)i + 36i + 72i^2 = -36sqrt(3) - 72sqrt(3)i + 36i - 72 Group the real and imaginary parts: = (-36sqrt(3) - 72) + (36 - 72sqrt(3))i Now, combine the numerator and denominator: (-36sqrt(3) - 72) + (36 - 72sqrt(3))i5 = -36sqrt(3) - 725 + 36 - 72sqrt(3)5i The final answer in standard form is *-36sqrt(3) - 725 + 36 - 72sqrt(3)5i. What's next?