If z= ( 1+2i 1-(1-i)^2 ), then arg(z) equal: - Vidyadip

MCQ

If $\text{z}=\Big(\frac{1+2\text{i}}{1-(1-\text{i})^2}\Big),$ then $\text{arg(z)}$ equal:

✓
$0$
B
$\frac{\pi}{2}$
C
$\pi$
D
none of these.

✓

Answer

Correct option: A.

$0$

$\text{z}=\frac{1+2\text{i}}{1-(1-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1+\text{i}^2-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1-1-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1+2\text{i}}$
$\Rightarrow\text{z}=1$
Since point $(1,0)$ lies on the positive direction of real axis, we have:
$\text{arg(z)} = 0$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Complex Numbers and Quadratic Equations→Complex Numbers and Quadratic Equations — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

If ${\log _{10}}2 = 0.30103,{\log _{10}}3 = 0.47712,$ the number of digits in ${3^{12}} \times {2^8} $ is

If the area of the triangle formed by the positive $x-$axis, the normal and the tangent to the circle $(x-2)^{2}+(y-3)^{2}=25$ at the point $(5,7)$ is $A$ then $24 A$ is equal to ...... .

Let $f(x) = \cos px + \sin x$ be periodic, then $p$ must be

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $ T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ Then $n ( S \cap T )$ is equal to $......$

If $4\text{x} + 3<6\text{x} + 7$, then$\text{ x}\in$

The sum of all possible values of $\theta \in[-\pi, 2 \pi]$, for which $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is purely imaginary, is equal to

The $...........$ is the difference between the greatest and the least value of the variate:

Let $A B C$ and $A B C^{\prime}$ be two non-congruent triangles with sides $A B=4$, $A C=A C^{\prime}=2 \sqrt{2}$ and angle $B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

Let $S$ be the sum of all solutions (in radians) of the equation $\sin ^{4} \theta+\cos ^{4} \theta-\sin \theta \cos \theta=0$ in $[0,4 \pi]$ Then $\frac{8 \mathrm{~S}}{\pi}$ is equal to ...... .

A common tangent to $9x^2 + 16y^2 = 144 ; y^2 - x + 4 = 0 \,\,\&\,\, x^2 + y^2 - 12x + 32 = 0$ is :