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1a. Find the modulus and argument of

Mathematics

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Here are the solutions to questions 1a and 1b.

1a. Find the modulus and argument of

For a complex number Z=x+yiZ = x + yi:

  • The modulus is ∣Z∣=x2+y2|Z| = \sqrt{x^2 + y^2}.
  • The argument is arg⁡(Z)=θ\arg(Z) = \theta, where tan⁡θ=yx\tan\theta = \frac{y}{x}, and θ\theta is chosen based on the quadrant of (x,y)(x, y).

i) Z=4+3iZ = 4 + 3i

  1. Identify the real and imaginary parts.

    Here, x=4x = 4 and y=3y = 3.

  2. Calculate the modulus.

    ∣Z∣=42+32=16+9=25=5|Z| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
  3. Calculate the argument.

    tan⁡θ=34\tan\theta = \frac{3}{4}

    Since x>0x > 0 and y>0y > 0, θ\theta is in the first quadrant.

    θ=arctan⁡(34)≈36.87∘ or 0.6435 radians\theta = \arctan\left(\frac{3}{4}\right) \approx 36.87^\circ \text{ or } 0.6435 \text{ radians}

    The modulus is 5\boxed{5} and the argument is arctan⁡(34)\boxed{\arctan\left(\frac{3}{4}\right)}.

    ii) Z=1+iZ = 1 + i

  4. Identify the real and imaginary parts.

    Here, x=1x = 1 and y=1y = 1.

  5. Calculate the modulus.

    ∣Z∣=12+12=1+1=2|Z| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}
  6. Calculate the argument.

    tan⁡θ=11=1\tan\theta = \frac{1}{1} = 1

    Since x>0x > 0 and y>0y > 0, θ\theta is in the first quadrant.

    θ=arctan⁡(1)=π4 or 45∘\theta = \arctan(1) = \frac{\pi}{4} \text{ or } 45^\circ

    The modulus is 2\boxed{\sqrt{2}} and the argument is π4\boxed{\frac{\pi}{4}}.

    iii) Z=sin⁡θ+icos⁡θZ = \sin\theta + i\cos\theta

  7. Identify the real and imaginary parts.

    Here, x=sin⁡θx = \sin\theta and y=cos⁡θy = \cos\theta.

  8. Calculate the modulus.

    ∣Z∣=(sin⁡θ)2+(cos⁡θ)2=sin⁡2θ+cos⁡2θ|Z| = \sqrt{(\sin\theta)^2 + (\cos\theta)^2} = \sqrt{\sin^2\theta + \cos^2\theta}

    Using the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1:

    ∣Z∣=1=1|Z| = \sqrt{1} = 1
  9. Calculate the argument.

    The standard polar form of a complex number is r(cos⁡ϕ+isin⁡ϕ)r(\cos\phi + i\sin\phi).
    We have Z=sin⁡θ+icos⁡θZ = \sin\theta + i\cos\theta.
    Using the trigonometric identities sin⁡θ=cos⁡(π2−θ)\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right) and cos⁡θ=sin⁡(π2−θ)\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right):

    Z=cos⁡(π2−θ)+isin⁡(π2−θ)Z = \cos\left(\frac{\pi}{2} - \theta\right) + i\sin\left(\frac{\pi}{2} - \theta\right)

    Comparing this to the polar form, the argument ϕ\phi is π2−θ\frac{\pi}{2} - \theta.
    The modulus is 1\boxed{1} and the argument is π2−θ\boxed{\frac{\pi}{2} - \theta}.

    1b. Evaluate

    The powers of ii follow a cycle: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1. This cycle repeats every 4 powers. To evaluate ini^n, find the remainder of nn when divided by 4.

    i) i20i^{20}

  10. Divide the exponent by 4.

    20÷4=5 with a remainder of 020 \div 4 = 5 \text{ with a remainder of } 0
  11. Use the remainder to find the value.

    Since the remainder is 0, i20i^{20} is equivalent to i4i^4.

    i20=i4=1i^{20} = i^4 = 1

    The value is 1\boxed{1}.

    ii) Z=−1+i2Z = -1 + i^2

  12. Substitute the value of i2i^2.

    We know i2=−1i^2 = -1.

    Z=−1+(−1)Z = -1 + (-1)
  13. Simplify the expression.

    Z=−1−1=−2Z = -1 - 1 = -2

    The value is −2\boxed{-2}.

    iii) i−4i^{-4}

  14. Rewrite the expression with a positive exponent.

    i−4=1i4i^{-4} = \frac{1}{i^4}
  15. Substitute the value of i4i^4.

    We know i4=1i^4 = 1.

    1i4=11=1\frac{1}{i^4} = \frac{1}{1} = 1

    The value is 1\boxed{1}.

    iv) i5i^5

  16. Divide the exponent by 4.

    5÷4=1 with a remainder of 15 \div 4 = 1 \text{ with a remainder of } 1
  17. Use the remainder to find the value.

    Since the remainder is 1, i5i^5 is equivalent to i1i^1.

    i5=i1=ii^5 = i^1 = i

    The value is i\boxed{i}.

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