Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [complex-numbers]

Questions involving complex numbers, that is numbers of the form $a+bi$ where $i^2=-1$ and $a,b\in\mathbb{R}$.

A complex number is a number in the form $z=a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit, or alternatively, $z=r\cdot e^{i\theta}$, with $r$ called the magnitude and $\theta$ called the argument.

The complex conjugate, $\overline z$, is $a-bi$ or $r\cdot e^{-i\theta}$.

Read more about complex numbers and their properties here.

19229 questions
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If $z_1=1-i$, $z_2=-2+4i$ and $a+ib=…$

If $z_1=1-i$, $z_2=-2+4i$ and $a+ib=\dfrac {z_1.z_2}{{z_1}^{'}}$ then the values of $a$ and $b$ are : $$a. 2,-4$$ $$b. 4, \dfrac {5}{2}$$ $$c. 1, 0$$ $$d. 4, \dfrac {1}{2}$$ My Attempt ; $$a+ib=\dfrac {(1-i)(-2+4i)}{(1+I)}$$ $$=\dfrac…
pi-π
  • 7,416
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Solve $a z - b \bar{z}^2 - c z^2 \bar{z} = 0$

How to solve the equation $$ a z - b \bar{z}^2 - c z^2 \bar{z} = 0 $$ for $z \in \mathbb{C}$ where $a, b, c \in \mathbb{C}$ are known coeffients. I want to solve for z. I started with substituting $z = r \, \mathrm{e}^{i \theta}$ to arrive at $$…
Timo
  • 189
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How to show that absolute value of a complex number is invariant under complex conjugation?

How to show that absolute value of a complex number is invariant under complex conjugation? my solution: Counterexample. $f(abs(1-i))= \overline{abs(1-i)}=\overline{1+i}=1-i \neq abs(1-i)$? is there something I am missing?
user2723
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How to prove $1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5\theta)}{\sin\theta} $?

I need help to prove that the following is true: $$1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5 \theta)}{\sin\theta}$$ I realize that I must evaluate the real part of this, but whatever I get I am not quite sure how…
Benjamin
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How can $|z-(i \bar w)| = |z-(-i \bar w)|$ where z and w are complex numbers?

How can $|z-(i \bar w)| = |z-(-i \bar w)|$ where z and w are complex numbers ? This is a step of the solution of a problem I'm doing and I don't understand why this is right. Isn't this like saying that $|z-a| = |z-(-a)| = |z+a| $ where z, a are…
Hema
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Prove if $|z|,|w|<1 $, then $|\frac{z-w}{1-zw}|<1 $ and if $|w|=|z|=1 $, then $\frac{z-w}{1-zw}\in{\Re}$

Prove: If $|z|,|w|<1$, then $\left|\frac{z-w}{1-z\bar{w}}\right|<1$. Not quite sure how to approach this.. I've tried squaring it and to do something from…
user565804
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Need help with complex numbers on an Argand diagram problem

Going through some complex number work for A-Level Further Maths and I have come across a question that I have had a crack at but the mark scheme is very limited so doesn't look at the method I tried to use, and I don't really understand how they…
user561475
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Arithmetic Progressions in Complex Variables

From Stein and Shakarchi's Complex Analysis book, Chapter 1 Exercise 22 asks the following: Let $\Bbb N=\{1,2,\ldots\}$ denote the set of positive integers. A subset $S\subseteq \Bbb N$ is said to be in arithmetic progression if…
anon271828
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if $|z|>1$ then there is $q \in \mathbb{Q} \cap [0, 2\pi)$ such that $Re(e^{iq}z)>1$

I want to show that if $z \in \mathbb{C}$ and $|z|>1$, then there is $q \in \mathbb{Q} \cap [0, 2\pi)$ such that $Re(e^{iq}z)>1$. Can you help me, please?. Thanks in advance.
user448150
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Find the number of roots of the equation $ z^4=\omega\overline{z} $ , for $ z \in \mathbb{C} $

Let $\omega=(\frac{3+4i}{3-4i})^5 $ Find the number of roots of the equation $ z^4=\omega\overline{z} $ , for $ z \in \mathbb{C} $ I am curious how to approach this type of problem, because after I computed this on Wolfram, I realised the solutions…
SADBOYS
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Conjugate of a complex number

Can you guys please tell that what's the conjugate of $i^i$? I tried to solve on the basis that multiplying conjugate with it's respective complex no. yields real no. So my try gives its answer as $i^{-i}$. Is it correct?
well...
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What is the infinite root of i?

I tried to calculate $$\sqrt{ ... {\sqrt{\sqrt i}}}$$ by saying $k =\sqrt{ ... {\sqrt{\sqrt i}}}$ , so the equation $$ \sqrt k = k$$ must be true. When we square each side of the equation, we get$$ k^2 - k = 0$$ so we get $\sqrt{ ... {\sqrt{\sqrt…
Jay Lee
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How do I find large powers of complex numbers using a+bi

I am always stuck on finding large powers of complex numbers, for example: Let z = $(\sqrt{3}+i)$. Find $z^{99}$ and express your answer in the form $"a+bi"$. Is there a trick in solving these? Can I use polar form in solving this?
peco
  • 309
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Finding a general formula for $\sqrt[n]{a+bi}$.

So $\sqrt[n]{a+bi}$ can be written as $$\exp\left(\dfrac{\ln(a+bi)}{n}\right).$$ However I don't know how to continue since I don't know a general rule for $\ln(a+bi)$.
Latin Wolf
  • 609
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Generalised Square of Sum of Modulus of Product of Complex Numbers

A cursory search of the site showed no answer I was looking for. Essentially I have a few questions. I know already that for $z_1, z_2 \in \mathbb{C}$, with arguments $\theta_1$ and $\theta_2$ respectively, we have $$ |z_1 + z_2|^2 = |z_1|^2 +…
Marko
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