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UPSC NDA Math Model Paper 9

Question : 68 of 120

Marks: +1, -0

Solution:

Given complex number Z = 3 + 3i.

Let rcosθ = 3 and rsinθ = 3

By squaring and adding, we get

r^2(\cos^2 \theta+\sin^2 \theta)=18

\therefore r=3\sqrt{2}

As we have taken

r \cos \theta=3

\Rightarrow \cos \theta=\frac{3}{r}=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}

⇒ θ = 45° (Since it is in first quadrant, there won’t be any changes in θ value)

So, on comparing with z = r (cosθ + i sinθ), we can write as

3\sqrt{2}

(cos 45° + i sin 45°)

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