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A complex number 
z_(1) has a magnitude 
|z_(1)|=5 and an angle 
theta_(1)=150^(@).
Express 
z_(1) in rectangular form, as 
z_(1)=a+bi.
Express 
a+bi in exact terms.

z_(1)=◻

A complex number z1 z_{1} has a magnitude z1=5 \left|z_{1}\right|=5 and an angle θ1=150 \theta_{1}=150^{\circ} .\newlineExpress z1 z_{1} in rectangular form, as z1=a+bi z_{1}=a+b i .\newlineExpress a+bi a+b i in exact terms.\newlinez1= z_{1}=\square

Full solution

Q. A complex number z1 z_{1} has a magnitude z1=5 \left|z_{1}\right|=5 and an angle θ1=150 \theta_{1}=150^{\circ} .\newlineExpress z1 z_{1} in rectangular form, as z1=a+bi z_{1}=a+b i .\newlineExpress a+bi a+b i in exact terms.\newlinez1= z_{1}=\square
  1. Rectangular Form Expression: To express a complex number in rectangular form, we use the polar form of the complex number, which is z=r(cos(θ)+isin(θ))z = r(\cos(\theta) + i\sin(\theta)), where rr is the magnitude and θ\theta is the angle.
  2. Substitute Values: Given z1=5\lvert z_{1} \rvert=5 and θ1=150\theta_{1}=150 degrees, we can substitute these values into the polar form equation.z1=5(cos(150 degrees)+isin(150 degrees))z_{1} = 5(\cos(150 \text{ degrees}) + i\sin(150 \text{ degrees}))
  3. Calculate Cosine and Sine: We need to calculate the cosine and sine of 150150 degrees. Since 150150 degrees is in the second quadrant, we know that cosine will be negative and sine will be positive.\newlinecos(150 degrees)=cos(180 degrees30 degrees)=cos(30 degrees)\cos(150 \text{ degrees}) = \cos(180 \text{ degrees} - 30 \text{ degrees}) = -\cos(30 \text{ degrees})\newlinesin(150 degrees)=sin(180 degrees30 degrees)=sin(30 degrees)\sin(150 \text{ degrees}) = \sin(180 \text{ degrees} - 30 \text{ degrees}) = \sin(30 \text{ degrees})
  4. Exact Trigonometric Values: We know the exact values for cos(30)\cos(30^\circ) and sin(30)\sin(30^\circ):cos(30)=32\cos(30^\circ) = \frac{\sqrt{3}}{2}sin(30)=12\sin(30^\circ) = \frac{1}{2}
  5. Substitute Exact Values: Substitute the exact values into the equation:\newlinez1=5(3/2+i(1/2))z_{1} = 5(-\sqrt{3}/2 + i*(1/2))
  6. Distribute Magnitude: Now, distribute the magnitude (55) to both the real and imaginary parts:\newlinez1=5×3/2+5×i×(1/2)z_{1} = 5 \times -\sqrt{3}/2 + 5 \times i\times(1/2)
  7. Simplify Expression: Simplify the expression: z1=532+(52)iz_{1} = -\frac{5\sqrt{3}}{2} + \left(\frac{5}{2}\right)i

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