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Math  /  Algebra

QuestionFind the exponential form for the complex number below z=(1+i)6(1+3i)6e(1+i)2z=\frac{(1+i)^{6}(1+\sqrt{3} i)^{6}}{e^{(1+i)^{2}}}

Studdy Solution

STEP 1

1. We need to express the complex number z z in exponential form.
2. The expression involves powers of complex numbers and the exponential function.
3. We will use Euler's formula and properties of exponents to simplify the expression.

STEP 2

1. Simplify the numerator (1+i)6(1+3i)6(1+i)^6 (1+\sqrt{3}i)^6.
2. Simplify the denominator e(1+i)2e^{(1+i)^2}.
3. Combine the results to express zz in exponential form.

STEP 3

Convert 1+i1+i into polar form. The modulus is:
∣1+i∣=12+12=2 |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}
The argument is:
arg⁡(1+i)=π4 \arg(1+i) = \frac{\pi}{4}
Thus, in polar form:
1+i=2(cos⁡π4+isin⁡π4) 1+i = \sqrt{2} \left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right)
Using Euler's formula:
1+i=2eiπ4 1+i = \sqrt{2} e^{i\frac{\pi}{4}}
Raise to the 6th power:
(1+i)6=(2)6ei6π4=8ei3π2 (1+i)^6 = (\sqrt{2})^6 e^{i\frac{6\pi}{4}} = 8 e^{i\frac{3\pi}{2}}

STEP 4

Convert 1+3i1+\sqrt{3}i into polar form. The modulus is:
∣1+3i∣=12+(3)2=4=2 |1+\sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2
The argument is:
arg⁡(1+3i)=π3 \arg(1+\sqrt{3}i) = \frac{\pi}{3}
Thus, in polar form:
1+3i=2(cos⁡π3+isin⁡π3) 1+\sqrt{3}i = 2 \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right)
Using Euler's formula:
1+3i=2eiπ3 1+\sqrt{3}i = 2 e^{i\frac{\pi}{3}}
Raise to the 6th power:
(1+3i)6=26ei6×π3=64ei2π=64 (1+\sqrt{3}i)^6 = 2^6 e^{i 6 \times \frac{\pi}{3}} = 64 e^{i 2\pi} = 64

STEP 5

Simplify the denominator e(1+i)2e^{(1+i)^2}. First, calculate (1+i)2(1+i)^2:
(1+i)2=12+2⋅1⋅i+i2=1+2i−1=2i (1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i
Thus, the denominator is:
e(1+i)2=e2i=cos⁡(2)+isin⁡(2) e^{(1+i)^2} = e^{2i} = \cos(2) + i\sin(2)

STEP 6

Combine the results to express zz in exponential form:
z=8ei3π2×64e2i=512ei3π2e2i z = \frac{8 e^{i\frac{3\pi}{2}} \times 64}{e^{2i}} = \frac{512 e^{i\frac{3\pi}{2}}}{e^{2i}}
Simplify the exponent:
z=512ei(3π2−2) z = 512 e^{i(\frac{3\pi}{2} - 2)}
This is the exponential form of zz.

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