Multiplication inverse of a number is number itself, then its initial value is

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4-1.Complex numbers

easy

The multiplication inverse of a number is the number itself, then its initial value is

$i$

$- 1$

$2$

$-i$

Solution

(b) We know that, multiplicative inverse of $x + iy$ is $\frac{1}{{x + iy}}$ According to question $x + iy = \frac{1}{{x + iy}}$ and only $(b)$ is satisfying such condition.

Standard 11

Mathematics

If $a = \cos \,\theta + i\,\sin \,\theta ,$ then $\frac{{1 + a}}{{1 – a}} = $

medium

If ${z_1}$ and ${z_2}$ are two complex numbers, then $|{z_1} + {z_2}|$ is

normal

The real part of ${(1 – \cos \theta + 2i\sin \theta )^{ – 1}}$is

medium

(IIT-1986)

If $(1 – i)^n = 2^n,$ then $n = $

medium

The least positive integer $n$ for which $\left( \frac{1 + i\sqrt 3 }{1 – i\sqrt 3 }\right)^n = 1,$ is?

hard

(JEE MAIN-2018)

The multiplication inverse of a number is the number itself, then its initial value is

Solution

Similar Questions

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