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Question: The mod -amp of \( - 1 - \sqrt 3 i\) is: A. \(cis\left( {\dfrac{{2\pi }}{3}} \right)\) B. \(2cis...

The mod -amp of −1−3i - 1 - \sqrt 3 i is:
A. cis(2π3)cis\left( {\dfrac{{2\pi }}{3}} \right)
B. 2cis(−2π3)2cis\left( {\dfrac{{ - 2\pi }}{3}} \right)
C. cis(−2π3)cis\left( {\dfrac{{ - 2\pi }}{3}} \right)
D. 2cis(2π3)2cis\left( {\dfrac{{2\pi }}{3}} \right)

Explanation

Solution

Suppose the complex number is z=a+ib=−1−3z = a + ib = - 1 - \sqrt 3 , where i denotes iota and the modulus of z is represented by ∣z∣=a2+(b)2=1+(3)2=4=2\left| z \right| = \sqrt {{a^2} + {{\left( b \right)}^2}} = \sqrt {1 + {{\left( {\sqrt 3 } \right)}^2}} = \sqrt 4 = 2 and argument of z is arg⁡(z)=tan⁡−1ba=−2π3\arg \left( z \right) = {\tan ^{ - 1}}\dfrac{b}{a} = \dfrac{{ - 2\pi }}{3} . Here the imaginary part is negative and the real part is also negative, so the angle is such that sin⁡θ\sin \theta and cos⁡θ\cos \theta will come out to be negative. Thus we can use trial and error methods and check the four options one by one. Here we are taking b as the imaginary part and a as the real part of the given complex number.

Complete step by step solution: Where a=−1a = - 1 and b=−3b = - \sqrt 3
Next we are going to find modulus of the given complex number.
Now modulus of z, ∣z∣=a2+(b)2=1+(3)2=4=2\left| z \right| = \sqrt {{a^2} + {{\left( b \right)}^2}} = \sqrt {1 + {{\left( {\sqrt 3 } \right)}^2}} = \sqrt 4 = 2
Hence z=2(−12,−32i)z = 2(\dfrac{{ - 1}}{2}, - \dfrac{{\sqrt 3 }}{2}i)
z=2(cos⁡(−2π3)+isin⁡(−2π3)) z=2e(i−2π3) \begin{gathered} z = 2\left( {\cos \left( {\dfrac{{ - 2\pi }}{3}} \right) + i\sin \left( {\dfrac{{ - 2\pi }}{3}} \right)} \right) \\\ z = 2{e^{\left( {i\dfrac{{ - 2\pi }}{3}} \right)}} \\\ \end{gathered}
Hence ∣z∣=2\left| z \right| = 2 and arg(z)=−2π3arg\left( z \right) = \dfrac{{ - 2\pi }}{3}

Note: In this type of question students always confuse with the argument solving part. Please note that you must use the value of sin⁡θ\sin \theta and cos⁡θ\cos \theta by carefully watching the corresponding quadrant in which the angle lies and knowledge of trigonometry is also required, like angle is taken negative for clockwise sense and also for angle 2π3\dfrac{{2\pi }}{3} which lies in 2nd quadrant and −2π3\dfrac{{ - 2\pi }}{3} which lies in 3rd quadrant, where sin and cos both are found to be negative. Try to remember the values of sin and cos for other angles if you want to solve these questions quickly. Pay attention to questions from complex number chapters as they are intuitive to solve.